The linear momentum $p$ of a body of mass $5 \,kg$ varies with time $t$ as $p = 5t^2 + t + 5$. It follows that the body is moving with:

$A$ particle of mass $2/3 \, kg$ with velocity $v = -15 \, m/s$ at $t = -2 \, s$ is acted upon by a force $F = k - \beta t^2$. Here,$k = 8 \, N$ and $\beta = 2 \, N/s^2$. The motion is one-dimensional. Then,the speed at which the particle acceleration is zero again,is ........... $m/s$.

$A$ body of mass $5\, kg$ starts from the origin with an initial velocity $\vec{u} = 30\hat{i} + 40\hat{j}\,m/s$. If a constant force $\vec{F} = -(\hat{i} + 5\hat{j})\,N$ acts on the body,the time in which the $y$-component of the velocity becomes zero is ......... $s$.

$A$ force $\vec{F_1}$ of magnitude $4 \ N$ acts on an object of mass $1 \ kg$ at the origin in a direction $30^{\circ}$ above the positive $x$-axis. $A$ second force $\vec{F_2}$ of magnitude $4 \ N$ acts on the same object in the direction of the positive $y$-axis. The magnitude of the acceleration of the object is nearly: (in $m \ s^{-2}$)

Which physical quantity is represented by the first derivative of linear momentum with respect to time?

$A$ force acting on a body of mass $5 \ kg$ is $(4 \hat{i} - 2 \hat{j} + 3 \hat{k}) \ N$. If the body is initially at rest,then the magnitude of its velocity at the end of $10 \ s$ in $m/s$ will be: