Question

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A body travels 200 cm in the first 2 s and 220 cm in the next 5 s. Calculate the velocity at the end of the seventh second from the start. (Ans. 2.22 m/s)​

Answer 1

Answer:

Here is the answerThe displacement of the body in first 2 sec = 200cm Let the initial velocity = u(say) , At t = 0, x(0) = 0, v (0) = u (say), x (t) = 200cm, t = 2s x(t') = (200 + 220)cm = 420cm t' = (2 + 5)s = 7s If a is the uniform acceleration of the particle, then x(t) = x(0) + v(0)t + � at^2 200 = 0 + u*2 + 0.5*a ...

Explanation:

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Answer 2

Answer:

200=2u+2a

u+a=100....(1)

For case 2. .i.e. next 4 seconds, we have distance =420 cm and time =6 s

420=6u+18a

u+3a=70....(2)

Solving equation 1 and 2, we get,

u=115 cm/s and a=−15 cm/s

2

Now by

v=u+at

v=115−15×7

v=10 cm/s

Hence answer is A.