Question

A body starts from rest and acquires a velocity of 12m/s is 5 sec calculate acceleration and distance?

Answer 1

Answer:

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,

$v=u+at$

Substitute the value,

$12=0+a(5)$

$5a=12$

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

$a=2.4$

So, Acceleration is 2.4 m/s².

Distance formula is

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

Substitute the value,

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

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

$s=30$

So, The distance is 30 m.

Answer 2

Answer:

Step-by-step explanation:

Given :-

Initial velocity, u = 0 (As it starts from rest)

Final velocity, v = 12 m/s

Time taken, t = 5 seconds

To Find :-

Acceleration, a = ??

Distance moved, s = ??

Formula to be used :-

1st Equation of motion, v = u + at

2nd Equation of motion, s = ut + 1/2 × at²

Solution :-

Putting all the values, we get

v = u + at

⇒ 12 = 0 + a × 5

⇒ 12 = 5a

⇒ 12/5 = a

⇒ a = 2.4 m/s²

Now, Distance moved,

s = ut + 1/2 × at²

⇒ s = 0 × 5 + 1/2 × 2.4 × 5 × 5

⇒ s = 1/2 × 2.4 × 25

⇒ s = 30 m

Hence, acceleration is 2.4 m/s² and distance moved is 30 m.