Question

A body starts from rest and acquires a velocity of 12 m s-1 in
5 s. Calculate the acceleration and distance moved.
(H.P.S.S.C.E. 2010 S, 2010, 2001)[Ans. 2.4 m s-2,30 m )​

Answer 1

Answer:

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.

Answer 2

Answer:

2.4 m/s², 30 metres

Explanation:

Given:

Initial velcoity = u = 0 m/s

Final velocity = v = 12 m/s

Time = t =  5 seconds

To Find:

• Acceleration
• Distance covered

Acceleration = $\frac{v-u}{t}$

Acceleration = $\frac{12-0}{5}$

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

Acceleration = 2.4 m/s²

Now we can find distance using second equation of motion:

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

Substituting the values:

s=0×5+$\frac{1}{2} \times 2.4 \times 25$

s=0+30

s=30m

The distance covered by the body is equal to 30 metres