Question

A 1.20 × 10^3 -kilogram car is traveling east at 25 meters per second. The brakes are applied and the car is brought to rest in 5.00 seconds. Calculate the magnitude and the direction of the total impulse applied to the car to bring it to rest

Answer 1

Given data:

* The mass of the car is,

$m=1.2\times10^3\text{ kg}$

* The initial velocity of the car is,

$u=25\text{ m/s}$

* The time taken by the car is t = 5 seconds.

* The final velocity of the car is v = 0 m/s.

Solution:

The impulse applied to the car is,

$J=mv-mu$

Substituting the known values,

$\begin{gathered} J=1.2\times10^3\times0-1.2\times10^3\times25 \\ J=0-30\times10^3 \\ J=-30\times10^3\text{ kgm/s} \end{gathered}$

Thus, the magnitude of the impulse applied on the car is,

$30\times10^3\text{ kgm/s}$

As the car comes to rest after the impulse is applied on the car.

Thus, the direction of impulse is opposite to the direction of motion of the car.