Question

9. Two bodies A and B start from rest from the same point with
a uniform acceleration of 2 ms? If B starts one second
later, then the two bodies are separated, at the end of the
next second, by
(a) 1 m (b) 2 m (c) 3 m (d) 4 m
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Answer 1

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.

Answer 2

a t

A 0 2m/s^2 2sec.

B 0 2m/s^2 1sec.

$\sf S_{A} = ut + \frac{1}{2} \: \: {at}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \frac{1}{ \cancel{2}} \times \cancel{2} \times 4 = 4m$

$\sf \:S_{B} = 4t + \frac{1}{2} \: \: {at}^{2} = \frac{1}{2} \times 2 \times 1 = 1m$

$\sf \:S_{A} - S_{B} = 4 - 1 = 3m$


$\underline {\boxed{ \color{red} \sf(c) \: 3m}}$