Question

A body travels 200 cm in the first 2 seconds and 220 cm in the next 5 seconds. Calculate the velocity at the end of the seventh second from the start

Answer 1

Answer:

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Explanation:

let initial velocity = u & accleration is a then

distance travelled in 4 sec = S6

using , s=ut + at2/2

         S6 = 4u + a36/2

              =  6u+18a

distance covered in 2 sec = S2 = 2u + 4a/2

                                          =2u+2a = 200           (given)

                                      u+a = 100             ........................1

now distance covered in 2 to 6 sec is given 220cm

     S6-200 = 220

    6u + 18a = 420       

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

from 1& 2

a = -15 & u=115

velocity after 7 sec = V7

      V7 = u + at

          =115 - 15*7=10cm/sec

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/s

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