A car moving with a speed of 90kmh\(^{-1}\) was brought uniformly to rest by the application of the brakes in 10s. How far did the car travel after the brakes were applied?
A
125m
B
150m
C
250m
D
15km
Verified Correct Answer: A.
Physical Principles & Numerical Solution
speed = \(\frac{90 \times 10^3}{3600}\) = 25ms\(^{-1}\)
Acceleration = \(\frac{\text{velocity change}}{\text{time}}\) = \(\frac{25}{10}\)
: Acceleration = 2.5
Using S = Ut + \(\frac{1}{2}\) at\(^2\)
S = \(\frac{1}{2}\) X 2.5 X 10\(^2\) = 125m
Source: WAEC / JAMB / NECO Classroom Archive (Record #28212)