Question : A particle is moving with a velocity of v=(3+6t+9t2) cms-1. What is the displacement of the particle in the interval t=5 to t=8 seconds?
Doubt by Japjee
Solution :
v=(3+6t+9t2) cms-1
t1=5 sec
t2=8 sec
We know,
v=dx/dt
vdt=dx
dx=vdt
dx=(3+6t+9t2)dt
Integrating both side within proper limits
When t1=5 sec then x1=x1
When t2=8 sec then x2=x2
s=3{[(8)+(8)2+(8)3]-[(5)+(5)2+(5)3]}
s=3[(8+64+512)-(5+25+125)}
s=3[584-155]
s=3[429]
s=1287 cm
s=12.87 m
Hence, 12.87 m is the displacement covered by the particle in the interval t=5 to t=8 seconds.