Question

A particle moves along thexaxis according to the equationx= 2t+ 3t^2, wherexisin m andtis in s. Calculate its instantaneous velocity and acceleration at t= 3s

Answer 1

Given that equation is as follows

x=2t+3t^2 which can be written as x=3t^2+2t

As we already know that distinguishing the given function with respect to t,we get instantaneous velocity

here we can write as V(inst)=dx/dt

putting the value of x in above formula we conclude

=d(3t^2+2t)/dt

=6t+2 (using differentiation  rule )

hence V(inst)=6t+2

for time t=3s,the instantaneous velocity is

V(t)=6t+2

V(3)=6(3)+2

V(3)=20 m/s

instantaneous velocity for given equation is 20 m/s.

it is already known that

a=dv/dt

where a is acceleration

here v=V(t)=6t+2

replacing v with its value

we get a=d(6t+2)/dt

a=6 m/s^2 (by using differentiation rule)

acceleration for given equation is 6 m/s^2.