The velocity ( mi ⁡ \operatorname{mi} / h r hr ) of a drone flying in the air is given by v ( t ) = 12 + 4 t 2 v\left(t\right)=12+4t^2 for 0 ≤ t ≤ 4 0\le t\le4 hours. Let s ( 0 ) = 0 s\left(0\right)=0 . Determine s ( t ) s\left(t\right) for 0 ≤ t ≤ 4 0\le t\le4 .

s ( t ) = 4 t 3 3 + 12 t s\left(t\right)=\frac{4t^3}{3}+12t

s ( t ) = 4 t 2 3 + 12 t s\left(t\right)=\frac{4t^2}{3}+12t

s ( t ) = 8 t s\left(t\right)=8t

s ( t ) = 8 t 3 3 + 4 t s\left(t\right)=\frac{8t^3}{3}+4t^{}

At t = 0 t=0 , a car approaching a stop sign decelerates from a speed of 50 mi ⁡ \operatorname{mi} / h r hr according to the acceleration function a ( t ) = 4 t + 3 a\left(t\right)=4t+3 , where t ≥ 0 t\ge0 and is measured in hours. How far does the car travel between t = 0 t=0 and t = 0.1 t=0.1 h r hr ?

5.02 mi ⁡ 5.02\operatorname{mi}

0.02 mi ⁡ 0.02\operatorname{mi}

50.32 mi ⁡ 50.32\operatorname{mi}

0.32 mi ⁡ 0.32\operatorname{mi}

A particle moves along the x x -axis and its acceleration is given by a ( t ) = cos ⁡ π t a\left(t\right)=\cos\pi t . Find v ( t ) v\left(t\right) if v ( 0 ) = 0 v\left(0\right)=0

v ( t ) = 1 π sin ⁡ ( π t ) v\left(t\right)=\frac{1}{\pi}\sin\left(\pi t\right)

v ( t ) = − π sin ⁡ ( π t ) v\left(t\right)=-\pi\sin\left(\pi t\right)

v ( t ) = − 1 π sin ⁡ ( π t ) v\left(t\right)=-\frac{1}{\pi}\sin\left(\pi t\right)

v ( t ) = 1 π cos ⁡ ( π t ) − 1 v\left(t\right)=\frac{1}{\pi}\cos\left(\pi t\right)-1

A particle moves along the x x x -axis and its acceleration is given by a ( t ) = cos ⁡ π t a\left(t\right)=\cos\pi t a ( t ) = cos π t . Find s ( t ) s\left(t\right) if s ( 0 ) = 1 s\left(0\right)=1

s ( t ) = π 2 − cos ⁡ ( π t ) + 1 π 2 s\left(t\right)=\frac{\pi^2-\cos\left(\pi t\right)+1}{\pi^2}

s ( t ) = π 2 − cos ⁡ ( π t ) π 2 s\left(t\right)=\frac{\pi^2-\cos\left(\pi t\right)}{\pi^2}

s ( t ) = π 2 + sin ⁡ ( π t ) + 1 π 2 s\left(t\right)=\frac{\pi^2+\sin\left(\pi t\right)+1}{\pi^2}

s ( t ) = π 2 + sin ⁡ ( π t ) π 2 s\left(t\right)=\frac{\pi^2+\sin\left(\pi t\right)}{\pi^2}

A rock is thrown from a height of 2 ft ⁡ 2\operatorname{ft} with an initial speed of 25 ft ⁡ 25\operatorname{ft} / s s . Acceleration resulting from gravity is − 32 ft ⁡ -32\operatorname{ft} / s 2 s^2 . Find v ( t ) v\left(t\right)

v ( t ) = 25 − 32 t v\left(t\right)=25-32t

v ( t ) = − 32 t v\left(t\right)=-32t

v ( t ) = 25 − 16 t 2 v\left(t\right)=25-16t^2

v ( t ) = 16 t 2 v\left(t\right)=16t^2

A rock is thrown from a height of 2 ft ⁡ 2\operatorname{ft} 2 ft with an initial speed of 25 ft ⁡ 25\operatorname{ft} 25 ft / s s s . Acceleration resulting from gravity is − 32 ft ⁡ -32\operatorname{ft} − 32 ft / s 2 s^2 s 2 . Find s ( t ) s\left(t\right)

s ( t ) = − 16 t 2 + 25 t + 2 s\left(t\right)=-16t^2+25t+2

s ( t ) = − 16 t 2 + 25 t s\left(t\right)=-16t^2+25t

s ( t ) = − 64 t 2 + 25 t + 2 s\left(t\right)=-64t^2+25t+2

s ( t ) = 16 t 2 s\left(t\right)=16t^2

10. Physics Applications of Integrals

Kinematics

10. Physics Applications of Integrals

Kinematics: Videos & Practice Problems

Video Lessons Practice Worksheet

Topic summary

Understanding motion involves the relationships between position, velocity, and acceleration. To find displacement, integrate the velocity function over a time interval, while total distance requires the absolute value of the velocity function to avoid negative values. For position, integrate the velocity function and add the initial position. The equations used include $s_{0}^{+}$ for position and $v_{0}^{+}$ for velocity.

concept

Using The Velocity Function

Video duration:

10m

Using The Velocity Function Video Summary

In the study of motion, understanding the relationship between velocity, displacement, distance, and position is crucial. When dealing with motion problems, we often start with the velocity function, which is the derivative of the position function with respect to time. To find displacement over a specific time interval, we can use integration, as it serves as the reverse process of differentiation.

For example, if the velocity of an object is given by the function $ v(t) = 2t - 4 $, to find the displacement from time $ t = 0 $ to $ t = 5 $, we integrate the velocity function over that interval:

$$ \text{Displacement} = \int_{0}^{5} v(t) \, dt = \int_{0}^{5} (2t - 4) \, dt $$

Calculating the antiderivative, we find:

$$ \int (2t - 4) \, dt = t^2 - 4t $$

Evaluating this from 0 to 5 gives:

$$ [5^2 - 4(5)] - [0^2 - 4(0)] = 25 - 20 = 5 $$

This result indicates that the displacement is 5 units. Graphically, this displacement corresponds to the area under the velocity curve, where areas above the time axis contribute positively and areas below contribute negatively.

When calculating total distance, however, we must consider that distance cannot be negative. Thus, we take the absolute value of the velocity function. For the same velocity function, we need to determine where the function is negative to set up our integrals correctly. The function $ v(t) = 2t - 4 $ becomes zero at $ t = 2 $. Therefore, we split the integral into two parts:

$$ \text{Total Distance} = -\int_{0}^{2} (2t - 4) \, dt + \int_{2}^{5} (2t - 4) \, dt $$

Calculating these integrals separately, we find:

1. For $ \int_{0}^{2} (2t - 4) \, dt $:

$$ = [t^2 - 4t]_{0}^{2} = (2^2 - 4(2)) - (0 - 0) = 4 - 8 = -4 $$

2. For $ \int_{2}^{5} (2t - 4) \, dt $:

$$ = [t^2 - 4t]_{2}^{5} = (5^2 - 4(5)) - (2^2 - 4(2)) = (25 - 20) - (4 - 8) = 5 + 4 = 9 $$

Thus, the total distance is:

$$ \text{Total Distance} = -(-4) + 9 = 4 + 9 = 13 $$

To find the position function, we integrate the velocity function and add the initial position. If the initial position $ s(0) = 3 $, the position function can be derived as follows:

$$ s(t) = s(0) + \int_{0}^{t} v(x) \, dx = 3 + \int_{0}^{t} (2x - 4) \, dx $$

Calculating the integral gives:

$$ = 3 + [x^2 - 4x]_{0}^{t} = 3 + (t^2 - 4t) $$

Thus, the position function is:

$$ s(t) = t^2 - 4t + 3 $$

To find the position at a specific time, simply substitute the desired time into the position function. This comprehensive understanding of displacement, distance, and position in relation to velocity functions is essential for solving motion problems effectively.

Problem

The velocity ( $mi$ / $h r$ ) of a drone flying in the air is given by $v (t) = 12 + 4 t^{2}$ for $0 is less than or equal to t is less than or equal to 4$ hours. Let $s of 0 equals 0$ .
Determine $s (t)$ for $0 is less than or equal to t is less than or equal to 4$ .

$s (t) = \frac{4 t^{2}}{3} + 12 t$

$s of t equals 8 t$

$s (t) = \frac{4 t^{3}}{3} + 12 t$

$s (t) = \frac{8 t^{3}}{3} + 4 t^{}$

Problem

The velocity ( $\operatorname{mi}$ / $hr$ ) of a drone flying in the air is given by $v\left(t\right)=12+4t^2$ for $0\le t\le4$ hours. Let $s\left(0\right)=0$ .
How far does the drone travel during the first hour?

13.3 mi

133.3 mi

8.00 mi

6.67 mi

Problem

The velocity ( $\operatorname{mi}$ / $h r$ ) of a drone flying in the air is given by $v\left(t\right)=12+4t^2$ for $0\le t\le4$ hours. Let $s\left(0\right)=0$ .
How far has the drone traveled by the time it has reached $48 mi$ / $h r$ ?

24 mi

34.7 mi

48 mi

72 mi

Problem

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v of t equals 2 cosine t$ for $0 is less than or equal to t is less than or equal to 2 pi$ .
Find the particle's displacement on $open bracket 0 comma 5 pi over 4 close bracket$

$- \sqrt{2}$

$- \frac{\sqrt{2}}{2}$

$- 2 - \sqrt{2}$

$2 + \sqrt{2}$

Problem

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v\left(t\right)=2\cos t$ for $0\le t\le2\pi$ .
Find total distance traveled by the particle on $\left\lbrack0,\frac{5\pi}{4}\right\rbrack$ .

1.41

2.00

5.41

0.027

Problem

A particle travels along the 𝑥-axis and its velocity is the given graph of $v (t)$ .
Find displacement on $open bracket 0 comma 5 close bracket$

2.5

4.5

-2

6.5

Problem

A particle travels along the 𝑥-axis and its velocity is the given graph of $v\left(t\right)$ .
Find total distance on $\left\lbrack0,5\right\rbrack$

2.5

4.5

6.5

concept

Using The Acceleration Function

Video duration:

Using The Acceleration Function Video Summary

Understanding the relationship between position, velocity, and acceleration is crucial in solving motion problems using integrals. These three concepts are interconnected through differentiation and integration. Specifically, the derivative of position with respect to time gives velocity, while the derivative of velocity gives acceleration. Conversely, integration allows us to move in the opposite direction: integrating velocity yields position, and integrating acceleration yields velocity.

To illustrate this, consider a scenario where the acceleration of a particle moving along the x-axis is defined by a function of time, expressed in meters per second squared. To find the velocity function from the given acceleration, we apply integration. The velocity function, $ v(t) $, can be determined using the formula:

\[v(t) = v(0) + \int_0^t a(x) \, dx \]

Here, $ v(0) $ represents the initial velocity, and $ a(x) $ is the acceleration function, which we replace with a dummy variable $ x $ for integration. After integrating the acceleration function, we evaluate it at the upper limit $ t $ and add the initial velocity. This process is similar to finding position from velocity, where the position function $ s(t) $ is given by:

\[s(t) = s(0) + \int_0^t v(x) \, dx \]

In this case, $ s(0) $ is the initial position. By substituting the velocity function into this integral, we can derive the position function.

For example, if we have an acceleration function $ a(t) = 6t - 12 $, we first integrate to find the velocity function:

\[v(t) = v(0) + \int_0^t (6x - 12) \, dx \]

After performing the integration and applying the initial condition, we arrive at a polynomial expression for velocity. Next, we can find the position function by integrating the velocity function:

\[s(t) = s(0) + \int_0^t v(x) \, dx \]

By evaluating this integral, we obtain the position function as a polynomial in terms of $ t $. Finally, to find the position of the particle at a specific time, such as $ t = 8 $ seconds, we substitute this value into the position function.

In summary, the key strategy for solving motion problems involving acceleration, velocity, and position is to utilize integration effectively. By integrating the acceleration to find velocity and then integrating velocity to find position, we can analyze the motion of particles over time. This systematic approach not only simplifies the problem-solving process but also reinforces the fundamental relationships between these three critical concepts in physics.

Problem

At $t equals 0$ , a car approaching a stop sign decelerates from a speed of 50 $mi$ / $h r$ according to the acceleration function $a (t) = 4 t + 3$ , where $t is greater than or equal to 0$ and is measured in hours. How far does the car travel between $t equals 0$ and $t equals 0.1$ $h r$ ?

$0.02 mi$

$50.32 mi$

$5.02 mi$

$0.32 mi$

Problem

A particle moves along the $x$ -axis and its acceleration is given by $a (t) = \cos π t$ .
Find $v (t)$ if $v of 0 equals 0$

$v (t) = \frac{1}{π} \sin (π t)$

$v (t) = - π \sin (π t)$

$v (t) = - \frac{1}{π} \sin (π t)$

$v (t) = \frac{1}{π} \cos (π t) - 1$

Problem

A particle moves along the $x$ -axis and its acceleration is given by $a\left(t\right)=\cos\pi t$ .
Find $s (t)$ if $s of 0 equals 1$

$s (t) = \frac{π^{2} - \cos (π t)}{π^{2}}$

$s (t) = \frac{π^{2} + \sin (π t) + 1}{π^{2}}$

$s (t) = \frac{π^{2} + \sin (π t)}{π^{2}}$

$s (t) = \frac{π^{2} - \cos (π t) + 1}{π^{2}}$

Problem

A rock is thrown from a height of $2 ft$ with an initial speed of $25 ft$ / $s$ . Acceleration resulting from gravity is $negative 32 ft$ / $s^{2}$ .
Find $v (t)$

$v (t) = - 32 t$

$v (t) = 25 - 32 t$

$v (t) = 25 - 16 t^{2}$

$v of t equals 16 t squared$

Problem

A rock is thrown from a height of $2\operatorname{ft}$ with an initial speed of $25\operatorname{ft}$ / $s$ . Acceleration resulting from gravity is $-32\operatorname{ft}$ / $s^2$ .
Find $s (t)$

$s (t) = - 16 t^{2} + 25 t + 2$

$s (t) = - 16 t^{2} + 25 t$

$s (t) = - 64 t^{2} + 25 t + 2$

$s of t equals 16 t squared$

example

Using The Acceleration Function Example 1

Video duration:

Using The Acceleration Function Example 1 Video Summary

A particle's motion can be analyzed through its acceleration, velocity, and displacement. When given an acceleration function, the first step is to derive the velocity function. The relationship between acceleration $ a(t) $ and velocity $ v(t) $ is established through integration. Specifically, the velocity function can be expressed as:

\[ v(t) = v(0) + \int_{0}^{t} a(x) \, dx \]

In this case, if the initial velocity $ v(0) $ is 6 meters per second, and the acceleration function is $ a(x) = 2x - 1 $, we can find the velocity function by integrating the acceleration:

\[ v(t) = 6 + \int_{0}^{t} (2x - 1) \, dx \]

Calculating the integral, we find:

\[ v(t) = 6 + \left[ x^2 - x \right]_{0}^{t} = 6 + (t^2 - t) \]

Thus, the velocity function is:

\[ v(t) = t^2 - t + 6 \]

Next, to find the displacement of the particle during the first five seconds, we integrate the velocity function over the interval from 0 to 5 seconds:

\[ s = \int_{0}^{5} v(t) \, dt = \int_{0}^{5} (t^2 - t + 6) \, dt \]

Calculating this integral gives:

\[ s = \left[ \frac{t^3}{3} - \frac{t^2}{2} + 6t \right]_{0}^{5} \]

Evaluating at the bounds, we find:

\[ s = \frac{125}{3} - \frac{25}{2} + 30 \approx 59.17 \text{ meters} \]

Finally, to determine the total distance traveled by the particle, we need to consider the absolute value of the velocity function. Since the velocity function $ v(t) = t^2 - t + 6 $ does not yield negative values for $ t \geq 0 $, we can integrate the same function without the absolute value:

\[ d = \int_{0}^{5} |v(t)| \, dt = \int_{0}^{5} (t^2 - t + 6) \, dt \]

Thus, the distance traveled is also:

\[ d \approx 59.17 \text{ meters} \]

In summary, the relationships between acceleration, velocity, displacement, and distance are interconnected through integration, allowing us to analyze the motion of the particle effectively. The displacement and distance traveled during the first five seconds are both approximately 59.17 meters, highlighting that the particle did not change direction during this interval.

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Here’s what students ask on this topic:

How do you find displacement using integrals in kinematics?

To find displacement in kinematics, you integrate the velocity function over a specified time interval. Displacement is essentially the change in position, and since velocity is the derivative of position with respect to time, integrating velocity gives you the position change. Mathematically, if $v (t)$ is the velocity function, the displacement $d$ over the interval $[a, b]$ is given by the integral: $d_{ab} = \int_{a, b} v (t) dt$ . This integral calculates the net area under the velocity-time curve, representing the total displacement.

What is the difference between displacement and total distance in kinematics?

In kinematics, displacement and total distance are related but distinct concepts. Displacement is the net change in position of an object and can be positive, negative, or zero. It is calculated by integrating the velocity function over a time interval. Total distance, on the other hand, is the sum of all distances traveled, regardless of direction, and is always a non-negative value. To find total distance, you integrate the absolute value of the velocity function over the interval. This ensures that any negative values, which indicate a change in direction, are accounted for as positive contributions to the total distance.

How do you find the position function from a velocity function in kinematics?

To find the position function from a velocity function in kinematics, you integrate the velocity function with respect to time and add the initial position. If $v (t)$ is the velocity function, the position function $s (t)$ is given by: $s (t) = s (t) + \int_{t, t} v (x) dx$ , where $s (t)$ is the initial position. This integration process provides a function that describes the position of the object at any time $t$ .

How can you find the velocity function from an acceleration function in kinematics?

To find the velocity function from an acceleration function in kinematics, you integrate the acceleration function with respect to time and add the initial velocity. If $a (t)$ is the acceleration function, the velocity function $v (t)$ is given by: $v (t) = v (t) + \int_{t, t} a (x) dx$ , where $v (t)$ is the initial velocity. This integration provides a function that describes the velocity of the object at any time $t$ .

What is the role of initial conditions in solving kinematics problems using integrals?

Initial conditions play a crucial role in solving kinematics problems using integrals. They provide the necessary constants needed to determine the specific solution to a problem. When integrating to find velocity from acceleration or position from velocity, the initial conditions, such as initial velocity or initial position, are added to the integral result. This ensures that the solution accurately reflects the motion of the object from the specified starting point. Without these initial conditions, the solution would be incomplete, as integration introduces an arbitrary constant that must be resolved using known values at a specific time.

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Kinematics: Videos & Practice Problems

Using The Velocity Function

Using The Velocity Function Video Summary

The velocity (mi\operatorname{mi}/hrhr) of a drone flying in the air is given by v(t)=12+4t2v\left(t\right)=12+4t^2 for 0≤t≤40\le t\le4 hours. Let s(0)=0s\left(0\right)=0. Determine s(t)s\left(t\right) for 0≤t≤40\le t\le4.

The velocity (mi\operatorname{mi}mi/hrhrhr) of a drone flying in the air is given by v(t)=12+4t2v\left(t\right)=12+4t^2v(t)=12+4t2 for 0≤t≤40\le t\le40≤t≤4 hours. Let s(0)=0s\left(0\right)=0s(0)=0.How far does the drone travel during the first hour?

The velocity (mi\operatorname{mi}mi/hrhr) of a drone flying in the air is given by v(t)=12+4t2v\left(t\right)=12+4t^2v(t)=12+4t2 for 0≤t≤40\le t\le40≤t≤4 hours. Let s(0)=0s\left(0\right)=0s(0)=0.How far has the drone traveled by the time it has reached 48mi48\operatorname{mi}/hrhr?

Suppose that a particle travels along the xx-axis and its velocity is given by v(t)=2costv\left(t\right)=2\cos t for 0≤t≤2π0\le t\le2\pi.Find the particle's displacement on [0,5π4]\left\lbrack0,\frac{5\pi}{4}\right\rbrack

Suppose that a particle travels along the xxx-axis and its velocity is given by v(t)=2costv\left(t\right)=2\cos tv(t)=2cost for 0≤t≤2π0\le t\le2\pi0≤t≤2π.Find total distance traveled by the particle on [0,5π4]\left\lbrack0,\frac{5\pi}{4}\right\rbrack[0,45π].

A particle travels along the 𝑥-axis and its velocity is the given graph of v(t)v\left(t\right). Find displacement on [0,5]\left\lbrack0,5\right\rbrack

A particle travels along the 𝑥-axis and its velocity is the given graph of v(t)v\left(t\right)v(t). Find total distance on [0,5]\left\lbrack0,5\right\rbrack[0,5]

Using The Acceleration Function

Using The Acceleration Function Video Summary

At t=0t=0, a car approaching a stop sign decelerates from a speed of 50 mi\operatorname{mi}/hrhr according to the acceleration function a(t)=4t+3a\left(t\right)=4t+3, where t≥0t\ge0 and is measured in hours. How far does the car travel between t=0t=0 and t=0.1t=0.1 hrhr?

A particle moves along the xx-axis and its acceleration is given by a(t)=cosπta\left(t\right)=\cos\pi t. Find v(t)v\left(t\right) if v(0)=0v\left(0\right)=0

A particle moves along the xxx-axis and its acceleration is given by a(t)=cosπta\left(t\right)=\cos\pi ta(t)=cosπt. Find s(t)s\left(t\right) if s(0)=1s\left(0\right)=1

A rock is thrown from a height of 2ft2\operatorname{ft} with an initial speed of 25ft25\operatorname{ft}/ss. Acceleration resulting from gravity is −32ft-32\operatorname{ft}/s2s^2. Find v(t)v\left(t\right)

A rock is thrown from a height of 2ft2\operatorname{ft}2ft with an initial speed of 25ft25\operatorname{ft}25ft/sss. Acceleration resulting from gravity is −32ft-32\operatorname{ft}−32ft/s2s^2s2. Find s(t)s\left(t\right)

Using The Acceleration Function Example 1

Using The Acceleration Function Example 1 Video Summary

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Here’s what students ask on this topic:

How do you find displacement using integrals in kinematics?

What is the difference between displacement and total distance in kinematics?

How do you find the position function from a velocity function in kinematics?

How can you find the velocity function from an acceleration function in kinematics?

What is the role of initial conditions in solving kinematics problems using integrals?

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The velocity ( $mi$ / $h r$ ) of a drone flying in the air is given by $v (t) = 12 + 4 t^{2}$ for $0 is less than or equal to t is less than or equal to 4$ hours. Let $s of 0 equals 0$ .
Determine $s (t)$ for $0 is less than or equal to t is less than or equal to 4$ .

The velocity ( $\operatorname{mi}$ / $hr$ ) of a drone flying in the air is given by $v\left(t\right)=12+4t^2$ for $0\le t\le4$ hours. Let $s\left(0\right)=0$ .
How far does the drone travel during the first hour?

The velocity ( $\operatorname{mi}$ / $h r$ ) of a drone flying in the air is given by $v\left(t\right)=12+4t^2$ for $0\le t\le4$ hours. Let $s\left(0\right)=0$ .
How far has the drone traveled by the time it has reached $48 mi$ / $h r$ ?

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v of t equals 2 cosine t$ for $0 is less than or equal to t is less than or equal to 2 pi$ .
Find the particle's displacement on $open bracket 0 comma 5 pi over 4 close bracket$

Suppose that a particle travels along the $x$ -axis and its velocity is given by $v\left(t\right)=2\cos t$ for $0\le t\le2\pi$ .
Find total distance traveled by the particle on $\left\lbrack0,\frac{5\pi}{4}\right\rbrack$ .

A particle travels along the 𝑥-axis and its velocity is the given graph of $v (t)$ .
Find displacement on $open bracket 0 comma 5 close bracket$

A particle travels along the 𝑥-axis and its velocity is the given graph of $v\left(t\right)$ .
Find total distance on $\left\lbrack0,5\right\rbrack$

At $t equals 0$ , a car approaching a stop sign decelerates from a speed of 50 $mi$ / $h r$ according to the acceleration function $a (t) = 4 t + 3$ , where $t is greater than or equal to 0$ and is measured in hours. How far does the car travel between $t equals 0$ and $t equals 0.1$ $h r$ ?

A particle moves along the $x$ -axis and its acceleration is given by $a (t) = \cos π t$ .
Find $v (t)$ if $v of 0 equals 0$

A particle moves along the $x$ -axis and its acceleration is given by $a\left(t\right)=\cos\pi t$ .
Find $s (t)$ if $s of 0 equals 1$

A rock is thrown from a height of $2 ft$ with an initial speed of $25 ft$ / $s$ . Acceleration resulting from gravity is $negative 32 ft$ / $s^{2}$ .
Find $v (t)$

A rock is thrown from a height of $2\operatorname{ft}$ with an initial speed of $25\operatorname{ft}$ / $s$ . Acceleration resulting from gravity is $-32\operatorname{ft}$ / $s^2$ .
Find $s (t)$