# Question 37707

Mar 29, 2017

$\left(a\right) : \text{ Appro. "3.47 sec., (b) :" Appro. } 434.02 m .$

#### Explanation:

Let, ${v}_{0} \mathmr{and} {v}_{t}$ be the Initial velocity and velocity after

time $t .$

Also, let $a$ be the Accelearation, and $s ,$ the Distance.

The following well-known eqns. of motion will be used to solve the

Problem.

$\left({M}_{1}\right) : {v}_{t} = {v}_{0} + a t , \mathmr{and} , \left({M}_{2}\right) : {v}_{t}^{2} = {v}_{0}^{2} + 2 a s .$

Part (a) :

v_0=100 (km)/(hr)=(100*1000)/3600 m/(sec)=250/9 m/sec, v_t=0, t=?.

Noting that, $a = - 8 \frac{m}{\sec} ^ 2 , \text{ we have, by, } \left({M}_{1}\right) , 0 = \frac{250}{9} - 8 t ,$

$\therefore t = \frac{250}{8 \cdot 9} = \frac{125}{36} \approx 3.47 \sec .$

Thus, the car will stop after appro. $3.47 \sec .$ after applying the brakes.

Part (b) :

In $\left({M}_{2}\right) ,$ we have, v_t=0, v_0=250/9 m/(sec.), a=-8 m/(sec)^2, s=?#

$\therefore 0 = {\left(\frac{250}{9}\right)}^{2} - 16 s \Rightarrow s = {\left(250\right)}^{2} / \left(9 \cdot 16\right) \approx 434.02 m .$

Enjoy Maths.!