The displacement x of a particle depend on ti | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Given, $x=\alpha t^2-\beta t^3$
Particle will return to its starting point when, $x =\alpha t ^2-\beta t ^3=0$
or,
$t=\frac{\alpha}{\beta}$
Veloci

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

Given, $x=\alpha t^2-\beta t^3$
Particle will return to its starting point when, $x =\alpha t ^2-\beta t ^3=0$
or,
$t=\frac{\alpha}{\beta}$
Velocity $= v =\frac{ dx }{ dt }=2 t \alpha-3 \beta t ^2 \ldots \ldots (1)$
at $t =0, v =0$ so the initial velocity zero.
Acceleration $= a =\frac{ dv }{ dt }=2 \alpha-6 t \beta \ldots \ldots(2)$
at $t =0, a =2 \alpha$ so initial acceleration does not zero.
The particle will come to rest when, 2 t $\alpha-3 \beta t ^2=0$ from $(1)$
or, $t=\frac{2 \alpha}{3 \beta}$
$At , t =\frac{\alpha}{3 \beta}, a =2 \alpha-6 \beta\left(\frac{\alpha}{3 \beta}\right)=0$
so net force $= ma =0$, thus no net force act on the particle when $t =\frac{\alpha}{3 \beta}$

The displacement $x$ of a particle depend on time $t$ as $x = \alpha {t^{^2}} - \beta {t^3}$

Similar Questions

The position of a particle moving along $x$-axis given by $x=\left(-2 t^3+3 t^2+5\right) m$. The acceleration of particle at the instant its velocity becomes zero is ....... $m / s ^2$

A particle moves in east direction with $15 \,m/sec$. for $2\, sec$ then moves northward with $5\, m/sec$. for $8 \,sec$. then average velocity of the particle is

A particle moves along a straight line in such a way that it’s acceleration is increasing at the rate of $2 m/s^3$. It’s initial acceleration and velocity were $0,$ the distance covered by it in $t = 3$ second is ........ $m$

What do you mean by term relative velocity ?