Question

An. object at. rest. gains. An. awarage. velocity. of. 12 m/sec. what. Will. be. it's. velocity at end of 4 m​

Answer 1

Acceleration is 2.4 m/s².

The distance is 30 m.

Step-by-step explanation:

Given : A body starts from rest and acquires a velocity of 12 m/s is 5 sec.

To find : Calculate acceleration and distance?

Solution :

We have given,

Initial velocity u=0 m/s

Final velocity v= 12 m/s

Time taken t=5 sec.

Applying law of motion,

v=u+atv=u+at

Substitute the value,

12=0+a(5)12=0+a(5)

5a=125a=12

a=\frac{12}{5}a=

5

12

a=2.4a=2.4

So, Acceleration is 2.4 m/s².

Distance formula is

s=ut+\frac{1}{2}at^2s=ut+

2

1

at

2

Substitute the value,

s=(0)(5)+\frac{1}{2}(2.4)(5)^2s=(0)(5)+

2

1

(2.4)(5)

2

s=\frac{1}{2}\times (2.4)\times 25s=

2

1

×(2.4)×25

s=30s=30

So, The distance is 30 m.

Answer 2

Step-by-step explanation:

object at. rest. gains. An. awarage. velocity. of. 12 m/sec. what. Will. be. it's. velocity at end of 4 m