Question

Two bodies a and b starts from rest from the same point with a uniform acceleration of 2 ms-2 if b starts one second later then the two bodies are separated at the end of next second by

Answer 1

B starts 1sec later than A.

So, at the end of next sec total travel time of A & B will be 2 sec & 1 sec respectively.

We have to find the dist of two bodies with their respective travel times.

DIST OF A:

S=ut + 1/2 at²

S=0 + 1/2 (2 * 2²) => S= 4m

DIST OF B:

S=0 + 1/2 (2 * 1²) => S= 1m

SEPARATION=DIST OF A - DIST OF B

Separation = 4 - 1 => 3m..

ALTERNATIVE:

Find the final velocity of B.

v=u+ at

v=0 + 2*1 => 2m/s

Now, imagine A will travel till the end of next sec with the above mentioned velocity as its initial velocity and the dist covered will be the separation gap.

Therefore,

S=ut+ 1/2 at²

S=(2 * 1) + 1/2 (2 * 1²) => 3m..

Answer 2

Explanation:

GIVEN :-

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

B stars one second later than A.

TO FIND :-

The distance between their distance.

SOLUTION :-

✮ For body A,

Initial velocity , u = 0 m/s . [ body starts from rest ]

Acceleration , a = 2 m/s².

Time , t = 2 seconds. [ say ]

✮ By using 2nd equation of motion,

➳ s = ut + 1/2 at²

➳ s = 0 × 2 + 1/2 × 2 × (2)²

➳ s = 0 + 1/2 × 2 × 4

➳ s = 0 + 1/2 × 8

➳ s = 0 + 4

➳ s = 4 m.

✮ For body B,

Initial velocity , u = 0 m/s . [ body starts from rest ]

Acceleration , a = 2 m/s².

Time , t = 1 seconds. [ say ]

✮By using 2nd equation of motion,

➠ s = ut + 1/2 at²

➠ s = 0 × 1 + 1/2 × 2 × (1)²

➠ s = 0 + 1/2 × 2 × 1

➠ s = 0 + 1/2 × 2

➠ s = 0 + 1

➠ s = 1 m.

✮ Difference between their distance,

➳ Distance of A - Distance of B

➳ 4 m - 1m.

➳ 3 m.