Question

Acceleration of a particle varies with time as a =
12t2 m/s2. If the particle starts its motion from
rest, then its displacement in initial 2 seconds
will be​ in

a.4m
b.64m
c.24m
d.16m​

Answer 1

Answer:

Acceleration of a particle varies with time as a =

12t2 m/s2. If the particle starts its motion from

rest, then its displacement in initial 2 seconds

will be in

Explanation:

d.16m

Answer 2

Given that acceleration of the particle varies with time as 12t².

Now,

$\displaystyle \sf \: a = \dfrac{dv}{dt} \\ \\ \longrightarrow \: \displaystyle \sf \: {12t}^{2} .dt = dv$

Integrating on both sides,

$\ \longrightarrow \: \displaystyle \sf \int {12t}^{2} .dt = \int dv \\ \\ \longrightarrow \sf \: v = 4 {t}^{3} \\ \\ \longrightarrow \sf \: \dfrac{dx}{dt} = {4t}^{3} \\ \\ \longrightarrow \sf \: dx = {4t}^{3} dt$

Integrating on both sides,

$\longrightarrow \displaystyle \int \sf \: dx = \int {4t}^{3} dt \\ \\ \longrightarrow \sf \: x = {t}^{4}$

Displacement during initial 2 seconds, i.e., from t = 0s to 2s.

$\longrightarrow \sf \: x = {2}^{4} - {0}^{4} \\ \\ \longrightarrow \boxed{ \boxed{\sf \: x = \: 16 \: m}}$

Option (d) is correct.