Question

A car travels with a uniform velocity of 72km/h. the driver applies the brakes and the car comes to rest with a uniform retardation in 10 seconds. find (i) the retardation (ii) velocity of the car after 3 seconds and (iii) the distance car travels after the braker are applied.​

Answer 1

Answer:if car stops therefore v1=0

Step-by-step explanation:

u=72x5/18=20m/s

Now, retardation= v-u/t

=-20/10=-2m/s^2

Velocity of car after 3 sec =v=u+at

=20+(-6)=14m/s

s=ut+1/2at^2

°s=200+100=300m

Answer 2

Answer:

(1) -2 m/$s^{2}$

(2) 14 m/s

(3) 100 m

Step-by-step explanation:

The initial velocity of the car u = 72km/h

                                                 = $\frac{5}{18}(72)$

                                             u = 20 m/s

If the car comes to rest after 10 seconds, the final velocity v = 0

We know that v = u + at → (1)

The retardation a, by the equation

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

$a=\frac{0-20}{10}$

a = -2 m/$s^{2}$

Therefore retardation after 10 seconds is 2 m/$s^{2}$

by the equation (1) , the velocity after 3 seconds is given by

v = 20 + (-2)3

v = 20 - 6

v = 14 m/s

Therefore the velocity after 3 seconds is 14m/s

We know that the distance s = ut + $\frac{1}{2}at^{2}$                                                            $s = 20(10) + \frac{1}{2} (-2)10^{2}$

s = 200 - 100

s = 100 m

Therefore the distance that the car travels after the brake is applied is

100 m