Question

A particle of mañs $s$moving along the $x$-axis experiences the net force$F_x=ct$, where $c$is a constant. The particle has velocity $v_{0x}$at$t=0$. Find an algebraic expression for the particle's velocity $v_x$ at a later time$t.$

Short answer

Thre expression of particle velocity at any time$t$is$v\left(t\right)=\frac{t^2}{2m}+v_{ta}$.

Step-by-step solution

Given.

Force on the particle, $F_x=ct.$

The velocity of the particle at a time$t=0,v0x.$

Formula used.

Force on an object is given by the equation$F=ma$.

Here, $m$is the mass, $a$is the acceleration.

Calculation.

Force on the particle is given as:

$F_X=ct\dots ..$(1)

Force on an object is given by the equation $F=ma$.........(2)

On comparing Eq. {1} and Eq. (2) $ma=ct$$a=\frac{a}{m}$

The velocity of a particle at a time is obtainedby integrating the above equation

$\begin{array}{rl}& \int adt=\int \frac{ct}{m}dt\\ & \int \frac{dv}{dt}dt=\frac{a^2}{2m}+C\cdots \cdots \left(3\right)\\ & v=\frac{a^2}{n}+C\end{array}$

Calculation

Here,

$C$is the arbitrary constant of integration

Apply the boundary condition that at time$t=0$the velocity of the particle is $v_{0x}$.

Plugging the values in the above equation

$\begin{array}{c}{v}_{bic}=\frac{ad}{m}+C\\ C=v_{0x}\end{array}$

Substitute the value of $C$in Eq. (3)

$v=\frac{a^2}{2m}+v0x$.

Conclusion

Thre expression of particle velocity at any time $t$is $v\left(t\right)=\frac{t^2}{2m}+v_{ta}$.