Question

a motorbike is traveling to the left with a speed of $27 \frac{\text{m}}{\text{s}}$ when the rider applies the brakes. the bike slows to a stop over $42 \text{ m}$ with constant acceleration.

what was the acceleration of the motorbike as it came to a stop?

kinematic equations
assume a coordinate system where rightward is positive. round your answer to two significant digits.

$\boxed{\quad} \frac{\text{m}}{\text{s}^2}$

Explanation:

Step1: Identify known values

Initial velocity \( v_0 = -27 \, \frac{\text{m}}{\text{s}} \) (negative because leftward is opposite to positive rightward), final velocity \( v = 0 \, \frac{\text{m}}{\text{s}} \), displacement \( x - x_0 = 42 \, \text{m} \). We use the kinematic equation \( v^2 = v_0^2 + 2a(x - x_0) \).

Step2: Rearrange the equation for acceleration \( a \)

From \( v^2 = v_0^2 + 2a(x - x_0) \), solve for \( a \):
\( a = \frac{v^2 - v_0^2}{2(x - x_0)} \)

Step3: Substitute the values

Substitute \( v = 0 \), \( v_0 = -27 \, \frac{\text{m}}{\text{s}} \), and \( x - x_0 = 42 \, \text{m} \):
\( a = \frac{0^2 - (-27)^2}{2 \times 42} = \frac{-729}{84} \approx -8.7 \, \frac{\text{m}}{\text{s}^2} \) (the negative sign indicates acceleration is to the right, decelerating the leftward motion)

Answer:

\(-8.7\) (or \(8.7\) with the understanding that direction is accounted for, but magnitude with sign as per coordinate system is \(-8.7\))