Question

a driver of a car traveling at 15.0m/s applies the brakes, causing a uniform acceleration of -2.0m/s^2. how long does it take the car to accelerate to a final speed of 10.0m/s? how far has the car moved during the breaking period?

Answer 1

V=V₀+at
at=V-V₀
t=(V-V₀)/a
t=(10-15)/(-2)=2.5 s
S=V₀t+at²/2
S=15*2.5-2*(2.5)²/2=31.25 meters

2.5 s and 31.25 m

Answer 2

Explanation:

Initial speed of the driver, u = 15 m/s

Acceleration of the driver, $a=-2\ m/s^2$

Final speed of the driver, v = 10 m/s

Let t is the time taken by the driver to accelerate to the final speed. Using first equation of kinematics as :

$t=\dfrac{v-u}{a}$

$t=\dfrac{10-15}{-2}$

t = 2.5 seconds

Let d is the distance the car moved during the breaking period. Using the third equation of kinematics as :

$v^2-u^2=2ad$

$d=\dfrac{v^2-u^2}{2a}$

$d=\dfrac{(10)^2-(15)^2}{2\times (-2)}$

d = 31.25 meters

Hence, this is the required solution.