Question

Velocity of a body is given as v = 3t2 + 2t + 1, then
displacement of the body in first 2 s will be (where
v is m/s and t is in s)
(1) 14m
(2) 10 m
(3) 8 m
(4) 20 m​

Answer 1

ut can be find using integration

so integrating v we get

s = 3t^3/3+2t^2/2+t

${t}^{3} + t {}^{2} + t$

=2^3+2^2+2

=14 m

Answer 2

Answer:

displacement of the body in first 2 s will be 14m.

Explanation:

since we know that $v=\frac{ds}{dt}$..................(1)

where ds is the displacement and dt is time taken by body to cover a specific displacement.

$vdt=ds$............from (1)

we have

$v=3t^{2}+2t+1$   so,

$(3t^{2}+2t+1)dt=ds$

integrating both sides...

$\int\ {3t^{2}+2t+1 } \, dt =\int\ {} \, ds$

we get,

$\frac{3t^{3} }{3}+\frac{2t^{2} }{2}+t=s$

$t^{3}+t^2+t=s$.................(2)

we need to find the displacement of the body in first 2sec.

put t=2 in (2) we get

$s=2^3+2^2+2$

s=14m.