Exams › NEET › Physics

The speed of a particle moving along a straight line is given by $v=t^2+3t-4$, where $v$ is in m/s and $t$ is in seconds. Find the time $t$ at which the particle will momentarily come to rest.

1. 3
2. 4
3. 2
4. 1

Correct answer: 1

Solution

Momentary rest means $v=0$. So solve $t^2+3t-4=0=(t+4)(t-1)$, giving $t=1$ or $t=-4$; physically, time must be non-negative, so $t=1$ s.

Related NEET Physics questions

• The radius $r$ and area $A = \pi r^2$ of a circle are both differentiable functions of time $t$. Write an equation relating $\frac{dA}{dt}$ to $\frac{dr}{dt}$.
• Find the roots of the quadratic equation \(2x^2 + 5x - 12 = 0\).
• A particle travels along a straight line with speed \(v\) (in m/s) given by \(v=t^2+3t-4\), where \(t\) is in seconds. At what time does the particle momentarily stop?
• Find $v(0)$, where $v(t)=3+2t$.
• $ \tan 15^\circ$ is equivalent to:
• The roots of the equation $2x^2+5x-12=0$ are:

⚔️ Practice NEET Physics free + battle 1v1 →