# The acceleration at t1 of a particle moving along the X-axis  is given by a(t)=20 t^3 + 6. At time t=0 the velocity of the particle is 7.What is the position of the particle at the time t?

Vinod Ramakrishnan Eswaran
41 Points
4 years ago
To find position we need to integrate acceleration two times to get distance(ie position)

a=20t3+6
v=∫a.dt+c  ( c is a constant)
$v=\int (a.dt)+c=\int (20t^{3}+6)dt+c$
$v=\frac{20t^{4}}{4}+6t+c$
Given at t=0 v=7
$7=\frac{20(0)^{4}}{4}+6(0)+c$
which implies c=7
The new equation is
$v=\frac{20(t)^{4}}{4}+6(t)+7$
$v=5(t)^{4}+6(t)+7$
Integrating again wrt t we get
$s=\int v.dt +c_{1}=\int (5(t)^{4}+6(t)+7) +c_{1}$
where c1 is again constant of integration
$s= (5\frac{(t)^{5}}{5}+\frac{6(t)^{2}}{2}+7t) +c_{1}$
$s= ((t)^{5}+3(t)^{2}+7t) +c_{1}$
During t=0 usually object covers 0 distance therefore
$0= ((0)^{5}+3(0)^{2}+7* (0)) +c_{1}$
which implies c1=0
Hence distance at time t is
$s= (t)^{5}+3(t)^{2}+7t$