In an emergency braking exercise, a student driver stops a car travelling at 83 km/h [W] in 4.0 s. What is the car’s acceleration (in m/s²)? Options: A) 21 [E] B) 5.8 [W] C) 21 [W] D) 5.8 [E]

What this question is really asking

You are finding the average acceleration while the car slows from an initial velocity to zero in a given time. The key parts are converting units correctly and keeping track of direction.

Convert the initial speed to m/s

Initial velocity: $83\ \text{km/h}$ [W]

$$83\ \text{km/h}=83\times\frac{1000\ \text{m}}{3600\ \text{s}}=23.06\ \text{m/s}$$ So $v_i=23.06\ \text{m/s}$ [W] and $v_f=0\ \text{m/s}$.

Compute acceleration from the velocity change

Use $$a=\frac{\Delta v}{\Delta t}=\frac{v_f-v_i}{t}$$

Let west be positive for the calculation: $v_i=+23.06\ \text{m/s}$, $v_f=0$.

$$a=\frac{0-23.06}{4.0}=-5.76\ \text{m/s}^2$$ The negative sign means the acceleration is opposite to [W], so it is toward [E].

Match to the answer choices

Magnitude $\approx 5.8\ \text{m/s}^2$, direction [E].

Correct choice: D) $5.8\ \text{m/s}^2$ [E].