Question

A body covers first 200m in 10 sec and next 220m in 10 sec withconstant acceleration. Find initial velocity, acceleration and velocity

after 20 sec of journey.​

Answer 1

Answer:

Explanation:

Correct answer -- (C) 10 cm/sec

Follow this method

Let 'u' be the initial velocity and 'a' the acceleration.

So we have the distance formula sut + 1/2 at^2

In first 2 seconds,

s = 200

Put the values in the formula. 2002u + 1/2 x 4 x a 200 = 2u + 2a 100 = u + a -----(1)

200 u x 2 + 1/2 x ax (2)^2

In next 4 seconds

Let the distance traveled be 'y' Time = 2+4= 6 sec

So according to the formula,

y = u x 6 + 1/2a x (6)^2 y = 6u + 1/2 x 36 x a

y = 6u + 18a ------(ii)

Now we know that 220 cm was traveled

in between 2 sec - 6 sec

y - 200 220

y = 420

We know y = 6u + 18a (from [ii])

So, 6u + 18a = 420

u + 3a = 70 ------(iii)

Equating (1) and (iii)

-2a = 30 a = -15 cm/s^2

u = 100 - (-15) u = 100 + 15 = 115 cm/sec

Now we know v = u + at

We have to find the "v" after 7th second

So v = 115 + (-15) x 7 v = 115 - 105 v = 10 cm/sec

Hope This Helps You!