An antelope moving with constant acceleration covers the distance between two points 70.070.0 m apart in 6.006.00 s. Its speed as it passes the second point is 15.015.0 m/s. What is its acceleration?

Ch 02: Motion Along a Straight Line

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 02: Motion Along a Straight Line

Problem 19b

Chapter 2, Problem 19b

An antelope moving with constant acceleration covers the distance between two points $70.0$ m apart in $6.00$ s. Its speed as it passes the second point is $15.0$ m/s. What is its acceleration?

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Identify the known values: initial distance (s) = 70.0 m, time (t) = 6.00 s, final velocity (v) = 15.0 m/s.

Use the kinematic equation for constant acceleration: $ v = u + at $, where $ v $ is the final velocity, $ u $ is the initial velocity, $ a $ is the acceleration, and $ t $ is the time.

Rearrange the equation to solve for the initial velocity $ u $: $ u = v - at $.

Use the second kinematic equation: $ s = ut + \frac{1}{2}at^2 $, where $ s $ is the distance covered.

Substitute the known values and the expression for $ u $ into the second equation to solve for acceleration $ a $.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Constant Acceleration

Constant acceleration refers to a situation where an object's velocity changes at a uniform rate over time. This means the acceleration value remains the same throughout the motion. In physics, this is often analyzed using kinematic equations, which relate displacement, initial velocity, final velocity, acceleration, and time.

Kinematic Equations

Kinematic equations are mathematical formulas used to describe the motion of objects under constant acceleration. They allow us to calculate unknown variables such as displacement, initial and final velocities, acceleration, and time. For this problem, the equation v = u + at and s = ut + 0.5at^2 are particularly useful, where v is final velocity, u is initial velocity, a is acceleration, t is time, and s is displacement.

Displacement and Velocity

Displacement is the vector quantity that refers to the change in position of an object, while velocity is the rate of change of displacement with respect to time. In this problem, the antelope's displacement is 70.0 m, and its final velocity at the second point is 15.0 m/s. Understanding these concepts is crucial for applying the kinematic equations to find the acceleration.

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A car's velocity as a function of time is given by $v_x(t) = α + βt^2$ , where $α = 3.00$ m/s and $β = 0.100$ m/s³. Draw $v_{x}$ - $t$ and $a_{x}$ - $t$ graphs for the car's motion between $t equals 0$ and $t equals 5.00$ s.

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An astronaut has left the International Space Station to test a new space scooter. Her partner measures the following velocity changes, each taking place in a $10$ -s interval. What are the magnitude, the algebraic sign, and the direction of the average acceleration in each interval? Assume that the positive direction is to the right.

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(b) At the beginning she is moving toward the left at $5.0$ m/s, and at the end she is moving toward the left at $15.0$ m/s.

(c) At the beginning she is moving toward the right at $15.0$ m/s, and at the end she is moving toward the left at $15.0$ m/s.

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A car's velocity as a function of time is given by $v_{x} (t) = α + β t^{2}$ , where $alpha equals 3.00$ m/s and $beta equals 0.100$ m/s³. Calculate the average acceleration for the time interval $t equals 0$ to $t equals 5.00$ s.

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