Question

10. A car accelerating uniformly from 15m/sec to 20 m/sec in 8 sec calculate
a) Its acceleration b) The distance covered by the car in that time​

Answer 1

Given:

Initial Speed (u) =15 m/s

Final Speed (v) =20 m/s

Time (t) =8 seconds

To Find:

a) Acceleration (a)=?

b) Distance=?

Solution:

v=u+at..... (Newton's First equation of motion)

20=15+a×8

20-15=8a

5=8a

a=5/8

a=0.625 m/s²

Therefore, Acceleration of the car is 0.625 m/s².

b) s=ut+1/2at².....(Newtons second equation of motion)

s=15×8+1/2×0.625×8×8

s=120+20

s=140m

Therefore,A car covers 140 m.

Answer 2

A car accelerates from 15m/s to 20 m/s in 8 sec.

Here, the initial velocity of the car is 15 m/s, final velocity of the car is 20 m/s and time is 8 sec.

We have to find the acceleration and the distance covered by the car.

a) For acceleration:

Using the First Equation Of Motion:

v = u + at

Substitute the known values,

20 = 15 + a(8)

20 - 15 = 8a

5 = 8a

0.6 = a

Therefore, the acceleration of the car is 0.6 m/s².

b) For distance:

Using the Second Equation Of Motion:

s = ut + 1/2 at²

Substitute the known values,

s = 15(8) + 1/2 × (0.6)(8)²

s = 120 + 0.3(64)

s = 120 + 19.2

s = 139.2

Therefore, the distance covered by the car is 139.2 m.