Question

a car and truck move with constant speed 20m/s the car is 10m behind the truck , the truck driver suddenly applies his brake ,causing the truck to accelerate at the constant rate of 2m/s. two seconds later the driver of the car applies his brake and just manages to avoid a rear end collision. determine the constant rate at which the car decelerate

Answer 1

Explanation:

truck parameters

u=20m\s

v=0

a=2m/s^2

t=10 sec. by 1st law of motion

cars parameters

u=20m\s

v=0

t=(10 -2) sec because breaks applied. after 2 sec

a=2.5m\s^2. by 1st law of motion

Answer 2

Answer:

3.33 m/s²

Explanation:

$v=u+at\\\Rightarrow 0=20-2t\\\Rightarrow t=\frac{-20}{-2}=10\ s$

Time taken by the truck to stop is 10 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=20\times 10+\frac{1}{2}\times -2\times 10^2\\\Rightarrow s=100\ m$

Distance traveled in the 10 seconds by the truck is 100 m

Distance traveled by the car in 2 seconds

Distance = Speed × Time

$\text{Distance}=20\times 2=40\ m$

Distance traveled by the car in 2 seconds is 40 m

60 m is remaining

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-20^2}{2\times 60}\\\Rightarrow a=-3.33\ m/s^2$

The constant rate of deceleration of the car is 3.33 m/s²