Question

A particle starts from rest with acceleration as a function of time given as a = 3t ms-2
Find the velocity and displacement of the particle 3 sec after the motion starts

Answer 1

Given that,

Acceleration $a=3t\ m/s^2$

Time = 3 sec

We need to calculate the velocity

Using formula of acceleration

$a=\dfrac{dv}{dt}$

$\dfrac{dv}{dt}=3t$

On integration of both side

$\int_{0}^{v}{dv}=\int_{0}^{t}{3t}dt$

$v=(\dfrac{3t^2}{2})_{0}^{3}+C$

$v=\dfrac{27}{2}+C$....(I)

At t= 0, v = 0

$0=0+C$

$C=0$

Put the value of C in equation (I)

$v=\dfrac{27}{2}$

$v = 13.5\ m/s$

We need to calculate the displacement

Using formula of velocity

$v=\dfrac{dx}{dt}$

$\dfrac{dx}{dt}=\dfrac{27}{2}$

$dx=\dfrac{27}{2}dt$

On integration of both side

$\int_{0}^{x}{dx}=\int_{0}^{t}{\dfrac{27}{2}dt}$

$x=(\dfrac{27}{2}t)_{0}^{3}+C'$

At t=0, x = 0 then, C = 0

$x=\dfrac{27}{2}\times3+0$

$x=40.5\ m$

Hence, The velocity and displacement of the particle is 13.5 m/s and 40.5 m.  

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Topic : velocity and displacement

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