Question

9.
The position x of a particle moving in one dimension, under the action of constant force is related to time t by
equation x = (t-3)^2
where x is in metre and t in sec. Calculate :
(a) The displacement of the particle when its velocity is zero;
(b) The work done by the force in the first 6 sec.​

Answer 1

Answer:

answer =0

Explanation:

Given equation can be rewritten as,

x=(t−3)

2

v=

dt

dx

=2(t−3)

v

f

=2(6−3)=6

v

i

=2(0−3)=−6

So, from work-energy theorem

W=ΔKE=

2

1

m[v

f

2

−v

i

2

]=

2

1

m[6

2

−(−6)

2

]=0

i.e., work done by the force in the first 6s is zero.