Question : Acceleration of a particle is given by a = 3t2-4t3. Then find velocity at t = 2 sec. If at t=0, u=2m/s also find position of the particle at t = 1 sec, if at t=0, x =0.
Doubt by Ankit
Solution :
We know,
a = dv/dt
adt=dv
dv=adt
Integrating both sides
v = ∫adt
v = ∫(3t²-4t³)dt
v= 3(t³/3)-4(t⁴/4) +c
where c is a constant of integration.
v=t³-t⁴+c — (1)
At t=0, v=u=2 m/s (Given)
2 = (0)³-(0)⁴+c
2 = c
c = 2
Putting in equation (1)
v=t³-t⁴+2
Now, Velocity at 2sec
v=(2)³-(2)⁴+2
v=8-16+2
v=10-16
v=-6 m/s
Also
v=dx/dt
vdt=dx
dx=vdt
integrating both sides
x=∫vdt
x=∫(t³-t⁴+2)dt [∵v=t³-t⁴+2]
x=t⁴/4-t⁵/5+2t+c — (2)
At t=0, x =0
0=(0)⁴/4-(0)⁵/5+2(0)+c
0=c
c=0
Putting c=0 in equation (2)
x=t⁴/4-t⁵/5+2t+0
x=t⁴/4-t⁵/5+2t
Now, the position at t=1sec
x=(1)⁴/4-(1)⁵/5+2(1)
x=1/4-1/5+2
x= (5-4+40)/20
x= (41/20)
x = 2.05 m