Question

A car starting from rest attains a speed of 54 km/hr in 2 s. Find

(a) acceleration produced

(b) distance covered during this time interval

​

Answer 1

Step-by-step explanation:

Given that,

Intial velocity u=0m/s

Final velocity v=54km/h=54×

18

5

=15m/s

Time t=2min=120s

Now, put the value in equation of motion

The acceleration is

v=u+at

a=

t

v−u

a=

120

15−0

a=o.125m/s

2

Now, again from equation of motion

The distance is

s=ut+

2

1

at

2

s=0+

2

1

×0.125×120×120

s=900m

Hence, the acceleration is 0.125m/s

2

and distance is 900m

Answer 2

The acceleration is found to be 7.5 m/$s^{2}$ and the distance it traversed is 15m.

Step-by-step explanation:

Given:

The initial velocity, u= 0 m/sec

Final velocity= 54km/hr= 15m/sec

Time taken to attain the velocity= 2 s

We need to find the acceleration produced by the car and the distance it covers.

Using the equation of motion:

v= u +at

where 'a' is acceleration and 't' is the time taken

Putting the values in above equation:

15= 0+ 2a

a= 15/2= 7.5 m/$s^{2}$

Now, using second equation  of motion:

s= ut+1/2 a$t^{2}$

s= 0x2+ 1/2 x 7.5 $(2)^{2}$

s= 0+ 1/2 x 7.5 x 4

s= 15 m

Result:

Thus, the acceleration is found to be 7.5 m/$s^{2}$ ad the distance it traversed is 15m.

(#SPJ3)