Question

Starting from rest a car acquires a velocity of 30km/h in 10 s and then brakes are applied ,it take 20s to stop.Calculate the distance travelled by the car.​

Answer 1

Answer:

2 km/h

I can't give the explanation

Answer 2

Given:

Starting from rest a car acquires a velocity of 30km/h in 10 s and then brakes are applied ,it take 20s to stop.

To find:

Distance travelled.

Conversion:

All units of speed has to be converted from km/hr to m/s.

30 km/hr = 30 × (5/18) = 8.33 m/s.

Calculation:

Let acceleration be a :

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

$= > \: a = \dfrac{8.33 - 0}{10}$

$= > \: a = \dfrac{8.33}{10}$

$= > \: a = 0.83 \: m {s}^{ - 2}$

Distance travelled in the first part of the journey when the car was accelerating :

$\therefore \: d1 = ut + \dfrac{1}{2} a {t}^{2}$

$= > \: d1 = 0 + \dfrac{1}{2} (0.83) {(10)}^{2}$

$= > \: d1 = \dfrac{1}{2} (0.83)(100)$

$\boxed{ = > \: d1 = 41.5 \: m}$

Now , let retardation be f:

$\therefore \: f = \dfrac{v - u}{t}$

$= > \: f = \dfrac{0 - 8.33}{20}$

$= > \: f = - 0.415 \: m {s}^{ - 2}$

Now , distance travelled in the second part of the journey when the car is retarding:

$\therefore \: d2 = ut + \dfrac{1}{2} f {t}^{2}$

$= > \: d2 = 8.33(20)+ \dfrac{1}{2} ( - 0.415) {(20)}^{2}$

$= > \: d2 = 166.6 - 83$

$\boxed{ = > \: d2 =83.6 \: m}$

So, total distance travelled :

$\therefore \: d = d1 + d2$

$= > \: d = 41.5 + 83.6$

$= > \: d = 125.1 \: m$

So, distance travelled is 125.1 metres.