Q1. A sports car starts at rest and accelerates uniformly to a speed of 40.0 m/s in 9.0 s. Calculate the distance the car travels during this time interval.

Background

Topic: Kinematics (Uniform Acceleration)

This question tests your understanding of motion with constant acceleration, specifically how to calculate the distance traveled when an object starts from rest and accelerates uniformly.

Key Terms and Formulas

• Uniform acceleration: Acceleration that does not change over time.

• Initial velocity ($v_0$): The velocity at the start (here, $v_0 = 0$ since the car starts from rest).

• Final velocity ($v$): The velocity at the end of the time interval ($v = 40.0$ m/s).

• Time ($t$): The duration of acceleration ($t = 9.0$ s).

• Distance ($x$): The total displacement during the time interval.

Key formulas:

• $v = v_0 + at$

• $x = v_0 t + \frac{1}{2} a t^2$

• Alternatively, $x = \frac{(v_0 + v)}{2} t$ (average velocity method)

Step-by-Step Guidance

1. Identify the known values: $v_0 = 0$ m/s (starts from rest), $v = 40.0$ m/s, $t = 9.0$ s.

2. First, find the acceleration ($a$) using the formula $v = v_0 + at$.

   Rearrange to solve for $a$:

   $a = \frac{v - v_0}{t}$

3. Plug in the known values to calculate $a$ (but do not compute the final value yet):

   $a = \frac{40.0\ \text{m/s} - 0}{9.0\ \text{s}}$

4. Now, use the kinematic equation for distance:

   $x = v_0 t + \frac{1}{2} a t^2$

   Since $v_0 = 0$, this simplifies to:

   $x = \frac{1}{2} a t^2$

5. Substitute the expression for $a$ and the value for $t$ into the equation for $x$ (but do not compute the final value):

   $x = \frac{1}{2} \left( \frac{40.0}{9.0} \right) (9.0)^2$

Try solving on your own before revealing the answer!