Question

an object starting from rest travels 20m in first 2 secs and 160m in next 4secs.what will be the velocity after 7secs fro the start

Answer 1

Answer:

70 m/s

Explanation:

For motion in first 2 s, we have

s=(1/2)at^2.

=> 20 = (1/2) * a * (2)²

=> 20 = (1/2) * a * 4

=> 20 = 2a

=> a = 10 m/s²

The velocity after 7 s from the start is :

v = at

  = (10)(7)

  = 70m/s

Hope it helps!

Answer 2

Given

Initial velocity =u=0m/s

distance traveled=S= 20m

time = t =2 sec

From second equation of motion :

S=ut+1/2at^2

20= 0*2+1/2a(2)^2

20=2a

a=10m/s^2

velocity at the end of 2 sec :

V=u+at

= 0+10*2

V=20m/s

After 4 sec it covers a distance of 160 m .

so,

160=20*4 +1/2(a^1)4^2

160=80+2a

8a^1=80

a^1=10m/s^2

It shows accelaration is uniform.

so velocity at the end of 7 sec from start :

V=u+at

=0+10*7

V=70m/s

Therefore, velocity after 7 sec from start is 70m/s