Question

Two bodies A and B starts from rest from the same point with a uniform acceleration of 2m/s2

. If B
starts one second later then the two bodies are separated, at the end of next second by :

1) 1m
2) 2m
3) 3m
4) 4m

Please give me step by step explanation....​

Answer 1

Answer:

Explanation:

Given,

Two bodies A and B start from rest from the same point with a uniform acceleration of 2 m/s².

And, B  starts one second later.

To Find,

Distance between them at the end.

Formula to be used,

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

Solution,

Distance covered by body A.

s = ut + 1/2 × at²

⇒ s = 0 + 1/2 × 2 × (2)²

⇒ s = 1/2 × 2 × 4

⇒ s = 1/2 × 8

⇒ s = 1/2 × 8

⇒ s = 4 m

Here distance coverd is 4 m,

Now, distance covered by body B,

s = ut + 1/2 × at²

⇒ s = 0 × 1 + 1/2 × 2 × (1)²

⇒ s = 1/2 × 2 × 1

⇒ s = 1/2 × 2

⇒ s = 1 m

Here, the distance covered is 1 m.

Now, the distance between them,

= Distance of (A - B)

= 4 - 1

= 3 m

Hence, the distance between the bodies is 3 m.

Answer 2

Given :-

Two bodies A and B satrts from rest from a same point.

Acceleration is 2 m/s².

B starts after one second later

To Find :-

Distance between them

Solution :-

The required disance

$\sf ut_1 + \dfrac{1}{2}a(t_1)^2 - ut_2+\dfrac{1}{2}a(t_2)^2$

$\sf u + \dfrac{1}{2} \times a \bigg(t_2^2-t_1^2\bigg)$

$\sf 0 + \dfrac{1}{2} \times 2 \bigg(2^2-1^2\bigg)$

$\sf 1 \times \bigg(4-1\bigg)$

$\sf 1 \times 3$

$\sf 3 m$