Basics of Dynamics

The velocity of a body at the end of $5 \mathrm{s}$ is $30 \mathrm{m} {\mathrm{s}}^{-1}$, at the end of 12 s is $58 \mathrm{m} {\mathrm{s}}^{-1}$ and at the end of 22 s, it is $98 \mathrm{m} {\mathrm{s}}^{-1}$. The body is moving with

A parachutist jumps out of an airplane and accelerates with gravity to a maximum velocity of $58.8 \mathrm{m}/\mathrm{s}$ in $6.00$ seconds. She then pulls the parachute cord and after a $4.00$ -second constant deceleration, descends at $10.0 \mathrm{m}/\mathrm{s}$ for $60.0$ seconds, reaching the ground. From what height did the parachutist jump?

The position $x$ of a particle with respect to time along $x$-axis is given by $x=9t^2-t^3$ where $x$ is in metres and $t$ in seconds. The position of this particle when it achieves maximum speed along the $+x$ direction will be at

The velocity of a body at the end of $5 \mathrm{s}$ is $30 \mathrm{m} {\mathrm{s}}^{-1}$, at the end of 12 s is $58 \mathrm{m} {\mathrm{s}}^{-1}$ and at the end of 22 s, it is $98 \mathrm{m} {\mathrm{s}}^{-1}$. The body is moving with

The displacement $x$ of a particle varies with time $t$ as $x=a{e}^{-\alpha t}+b{e}^{-\beta t}$ where $a, b, \alpha , \beta$ are positive constants. The velocity of the particle will

The motion of a particle is represented by the equation $\mathrm{S}=8{\mathrm{t}}^3-2{\mathrm{t}}^2+6\mathrm{t}+7$. Find the velocity of the particle at the end of $2$ seconds in $\mathrm{m}/\mathrm{s}$.

The displacement of a particle moving along a straight line is related to time as $\sqrt{x}=t-3$. The displacement of the particle when its velocity is zero is

The motion of a particle is given by the equation $S=\left(3t^3+7t^2+14t+8\right) \mathrm{m}$. The value of acceleration of the particle at $t=1$ sec is
 

The displacement of a particle is represented by the following equation : $x=3t^3+7t^2+5t+8$ where $x$ is in metre and $t$ in second. The acceleration of the particle at $t=1$ sec is :-

The displacement $y$ (in metres) of a body varies with time $t$ (in seconds) as $y=\frac{-2}{3}{t}^2+16t-12$. Then, the body will come to rest in

The displacement of a body from a reference point is given by $\sqrt{x}=2t-3,$ where $x$ is in metres and $t$ in seconds. This shows that the body:

If a particle's position is given by $x=4-12t+3t^2$ where $t$ is in the seconds and $x$ in meters. What is its velocity at $t=1s$? Whether the particle is moving in positive $x$ direction or negative $x$ direction?

The correct statement from the following is

The motion of a particle along a straight line is described by the function $x={\left(2t-3\right)}^2$ where $x$ is in metres and $t$ is in seconds. The acceleration of the particle at $t=2 \mathrm{s}$ is

The acceleration at the end of $2 \mathrm{s}$, of a particle whose motion is represented by the equation $S=4{\mathrm{t}}^3-8{\mathrm{t}}^2+5\mathrm{t}+4$ is ______

The displacement of a particle moving with uniform acceleration in time $t$ is given by $S=30t+5{\mathrm{t}}^2$, its initial velocity is _____

The horizontal and vertical displacements of a particle moving along a curved line are given by $x=5t$ and $y=\left(2t^2+t\right)$ metre . Time after which its velocity vector makes an angle of $45°$ with the horizontal is

A scooter going due east at $10 {\mathrm{ms}}^{-1}$  turns right through an angle of ${90}^o$. What is the change in velocity of scooter on taking turn if its speed remains constant?

The displacement of a body along X-axis depends on time as $\sqrt{x}=t+1$. Then the velocity of the body

The displacement of a body is given by $2s=g{t}^2$ where $g$ is a constant. The velocity of the body at any time $t$ is: