Question

a body covers 200cm in the first 2 seconds and 220 CM in next 4 seconds what is the velocity of the body at the end of 7 second

Answer 1

We are assuming that the body is under uniform acceleration here. Then, the total distance d travelled after the first t seconds is given by

d=ut+12at2d=ut+12at2

Where u is the initial velocity and a is the acceleration. We put the values of these quantities for the first 2 seconds:

200=2u+2a200=2u+2a

where u is in cm/s and a is in cm/s/s.

Then at the end of 6 seconds, the total distance travelled is 200+220 = 420 cm. So,

420=6u+18a420=6u+18a

Upon solving, this gives u = 115 cm/s and a = -15 cm/s/s.

So velocity at the end of 7 seconds is given by:

v=u+atv=u+at

= 115 - 105 = 10 cm/s

Answer 2

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

So we have the distance formula

s = ut + 1/2 at^2

In first 2 seconds,

s = 200

Put the values in the formula.

200 = u x 2 + 1/2 x a x (2)^2

200 = 2u + 1/2 x 4 x a

200 = 2u + 2a

100 = u + a -----(I)

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 (I) 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

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