Question

A car at rest accelerates uniformly to a speed of 144 km h-in 20 s. How much distance is covered by the car ?

(H.P.S.S.C.E. 2010 S)[Ans. 1,200 m)​

Answer 1

Answer:-

S = 400 m

Given :-

u = 0m/s.

v = 144 km/hr

=144 × 5/18

= 40 m/s

T = 20 s

To find :-

The distance covered by car.

Solution:-

Distance:- The length of the actual path covered by a body in a given time is called distance .

• It is a scalar quantity.

Let the distance covered by car be S.

Let first find acceleration produced by the car.

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

put the given value ,

→$a = \dfrac {40-0}{20}$

→$a = 2 m/s^2$

Now , the distance travelled by car is given by :-

$2aS = v^2 - u^2$

→$2 \times 2 \times S = 40^2 - 0^2$

→$4 \times S = 1600$

→$S = \dfrac{1600}{4}$

→$S = 400 m$

hence,

The distance covered by car will be 400 m .

Answer 2

Answer:

400 m .

Explanation:

Given :

Initial velocity ( u ) = 0 m / sec

Final velocity ( v ) = 144 km / hr

To convert km / hr into m / sec multiply it by 5 / 18 we get

v = 8 × 5 = 40 m / sec

Time ( t ) = 20 sec

First find the acceleration ( a )

From first equation of motion we have

v = u + a t

Put the value here we get

40 = 0 + 20 a

20 a = 40

a = 2 $m/sec^2$

Now we have second equation of motion

$\displaystyle s=ut+1/2at^2$

u = Initial velocity  

 a = Acceleration

 t = Time

s = distance

put the value here we get

$\displaystyle s=ut+1/2at^2$

u = Initial velocity

a = Acceleration

t = Time

s = distance

put the value here we get

$\displaystyle s=0\times20+1/2\times2\times(20)^2\\\\\displaystyle s=1/2\times2\times(20)^2\\\\\displaystyle s=(20)^2\\\\\displaystyle s=400 \ m$

Thus we get distance 400 m .