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The displacement x of a particle along a straight line at time t is given by $\mathrm{x}={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{t}+{\mathrm{a}}_2{\mathrm{t}}^2$. The acceleration of the particle is
(1)  ${\mathrm{a}}_0$
(2)  ${\mathrm{a}}_1$
(3)  $2{\mathrm{a}}_2$
(4)  ${\mathrm{a}}_2$

The relation between time and distance is $\mathrm{t}={\mathrm{\alpha x}}^2+\mathrm{\beta x}$, where $\mathrm{\alpha }$ and $\mathrm{\beta }$ are constants. The retardation is:
1.  $2{\mathrm{\alpha v}}^3$
2.  $2{\mathrm{\beta v}}^3$
3.  $2{\mathrm{\alpha \beta v}}^3$
4.  $2{\mathrm{\beta }}^2{\mathrm{v}}^3$

The displacement of a particle is given by $\mathrm{y}=\mathrm{a}+\mathrm{bt}+{\mathrm{ct}}^2-{\mathrm{dt}}^4$. The initial velocity and acceleration are respectively
(1)  b, -4d
(2)  -d, 2c
(3)  b, 2c
(4)  2c, -4d

The momentum is given by p=4t+1, the force at t=2s is
(1)  4 N
(2)  8 N
(3)  10 N
(4)  15 N

A particle moves along a straight line such that its displacement at any time t is given by $s = t^3 - 3t^{2 }+ 6t - 12$ metres. The velocity when the acceleration is zero is:
(A)  $3 {\mathrm{ms}}^{-1}$
(B)  $-12 {\mathrm{ms}}^{-1}$
(C)  $42 {\mathrm{ms}}^{-1}$
(D)  $-9 {\mathrm{ms}}^{-1}$

The position x of particle varies with time t as $\mathrm{x}={\mathrm{at}}^2-{\mathrm{bt}}^3$. The acceleration of the particle will be zero at time equal to
(1)  $\frac{\mathrm{a}}{\mathrm{b}}$
(2)  $\frac{2\mathrm{a}}{\mathrm{b}}$
(3)  $\frac{\mathrm{a}}{3\mathrm{b}}$
(4)  Zero

A body is moving according to the equation $\mathrm{x}=\mathrm{at}+{\mathrm{bt}}^2-{\mathrm{ct}}^3$ where x = displacement and a, b and c are constants. The acceleration of the body is
(1)  $\mathrm{a}+2\mathrm{bt}$
(2)  $2\mathrm{b}+6\mathrm{ct}$
(3)  $2\mathrm{b}-6\mathrm{ct}$
(4)  $3\mathrm{b}-6{\mathrm{ct}}^2$

A particle moves along X-axis in such a way that its X coordinate  varies with time t according to the equation $\mathrm{x}=\left(2-5\mathrm{t}+6{\mathrm{t}}^2\right)\mathrm{m}$. The initial velocity of the particle is
(1)  -5 m/s
(2)  6 m/s
(3)  3 m/s
(4)  3 m/s

If the momentum of a particle is given by P=(180-8t) kg m/s, then its force will be
(1)  Zero
(2)  8 N
(3)  -8 N
(4)  4 N

A particle moves along x-axis as $\mathrm{x}=4\left(\mathrm{t}-2\right)+\mathrm{a}{\left(\mathrm{t}-2\right)}^2$
Which of the following is true?
(1)  The initial velocity of particle is 4
(2)  The acceleration of particle is 2a
(3)  The particle is at origin at t=0
(4)  None of these

The maximum value of $7+6\mathrm{x}-9{\mathrm{x}}^2$ is
(A)  8
(B)  -8
(C)  4
(D)  -4

If $\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2-2\mathrm{x}+4$, then \(f(x)\) has:
1. a minimum at \(x=1\)
2. a maximum at \(x=1\)
3. no extreme point
4. no minimum

A particle is moving along the x-axis. The velocity v of this particle varies with its position x as $\mathrm{v}=\frac{1}{\mathrm{x}}$. Find the velocity of the particle as a function of time t given that at t=0, x=1 and v=$\frac{\mathrm{dx}}{\mathrm{dt}}$.
1. $\mathrm{v}=\sqrt{2\mathrm{t}+1}$
2. $\mathrm{v}=\frac{1}{\sqrt{2\mathrm{t}+1}}$
3. $\mathrm{v}=\frac{1}{\mathrm{t}}$
4. None of these

A particle moving along a straight line according to the law $\mathrm{x}=\mathrm{At}+{\mathrm{Bt}}^2+{\mathrm{Ct}}^3$, where x is its position measured from a fixed point on the line and t is the time elapsed till it reaches position x after starting from the fixed point. Here A, B and C are positive constants.
(1)  Its velocity at t=0 is A
(2)  Its acceleration at t=0 is B
(3)  Its velocity at t=0 is B
(4)  Its acceleration at t=0 is C

If the velocity of a particle moving on x-axis is given by $\mathrm{v}=3{\mathrm{t}}^2-12\mathrm{t}+6$. At which time is the acceleration of particle zero?
1. $2$ sec
2. $2+\sqrt{2}$ sec
3. $2-\sqrt{2}$ sec
4. zero

Momentum of a body moving in a straight line is $\mathrm{p}=\left({\mathrm{t}}^2+2\mathrm{t}+1\right) \mathrm{kg} \mathrm{m}/\mathrm{s}$. Find the force acting on a body at t=2 sec
(A)  6 N
(B)  8 N
(C)  4 N
(D)  2 N

A particle moves along straight line such that at time t its position from a fixed point O on the line is $\mathrm{x}=3{\mathrm{t}}^2-2$. The velocity of the particle when t=2 is:
(A)  $8 {\mathrm{ms}}^{-1}$
(B)  $4 {\mathrm{ms}}^{-1}$
(C)  $12 {\mathrm{ms}}^{-1}$
(D)  $0$

The coordinate of an object is given as a function of time by $\mathrm{x}=7\mathrm{t}-3{\mathrm{t}}^2$, where x is in meters and t is in seconds. Its average velocity over the interval from t=0 to t=4 is:
(1)  5 m/s
(2)  -5 m/s
(3)  11 m/s
(4)  -11 m/s

Coordinates of a moving particle are given by $\mathrm{x}={\mathrm{ct}}^2$ and $\mathrm{y}={\mathrm{bt}}^2$. The speed of the particle is given by
(1)  $2\mathrm{t} \left(\mathrm{c}+\mathrm{b}\right)$
(2)  $2\mathrm{t} \sqrt{{\mathrm{c}}^2-{\mathrm{b}}^2}$
(3)  $\mathrm{t} \sqrt{{\mathrm{c}}^2+{\mathrm{b}}^2}$
(4)  $2\mathrm{t} \sqrt{{\mathrm{c}}^2+{\mathrm{b}}^2}$

A particle moves along a straight line and its position as a function of time is given by$\mathrm{}$\(x= t^3-3t^2+3t+3\) then particle:

Momentum of a body moving in a straight line is $\mathrm{p}=\left({\mathrm{t}}^2+2\mathrm{t}+1\right) \mathrm{kg} \mathrm{m}/\mathrm{s}$. Force acting on a body at t=2 sec
(A)  6 N
(B)  8 N
(C)  4 N
(D)  2 N

A particle moves along a straight line such that at time t its position from a fixed point O on the line is $3{\mathrm{t}}^2-2$. The velocity of the particle when t=2 is:
(A)  8 ms^-1
(B)  4 ms^{-1
(C)  12 ms-1
(D)  0}

A particle moves in a straight line, according to the law $\mathrm{x}=4\mathrm{a}\left[\mathrm{t}+\mathrm{a} \sin \left(\frac{\mathrm{t}}{\mathrm{a}}\right) \right]$, where x is its position in meters, t in sec & a is some constants, then the velocity is zero at
(1)  $\mathrm{x}=4{\mathrm{a}}^2\mathrm{\pi } \mathrm{meters}$
(2)  $\mathrm{t}=\mathrm{\pi } \sec$
(3)  $\mathrm{t}=0 \sec$
(4)  none

If the distance 's' travelled by a body in time 't' is given by $\mathrm{s}=\frac{\mathrm{a}}{\mathrm{t}}+{\mathrm{bt}}^2$ then the acceleration equals
(1)  $\frac{2\mathrm{a}}{{\mathrm{t}}^3}+2\mathrm{b}$
(2)  $\frac{2\mathrm{s}}{{\mathrm{t}}^3}$
(3)  $2\mathrm{b}-\frac{2\mathrm{a}}{{\mathrm{t}}^3}$
(4)  $\frac{\mathrm{s}}{{\mathrm{t}}^2}$

A point moves in a straight line so that its displacement is x m at time t sec, given by ${\mathrm{x}}^2={\mathrm{t}}^2+1$. Its acceleration in $\mathrm{m}/{\mathrm{s}}^2$ at time 1 sec is:
(1)  $\frac{1}{\mathrm{x}}$
(2)  $\frac{1}{\mathrm{x}}-\frac{1}{{\mathrm{x}}^2}$
(3)  $\frac{1}{{\mathrm{x}}^3}$
(4)  $-\frac{{\mathrm{t}}^2}{{\mathrm{x}}^3}$

The velocity of a particle moving on the x-axis is given by $\mathrm{v}={\mathrm{x}}^2+\mathrm{x}$ where v is in m/s and x is in m. Find its acceleration in $\mathrm{m}/{\mathrm{s}}^2$ when passing through the point x=2m
(1)  0
(2)  5
(3)  11
(4)  30

A particle moves in space such that
$\mathrm{x}=2{\mathrm{t}}^3+3\mathrm{t}+4 ; \mathrm{y}={\mathrm{t}}^2+4\mathrm{t}-1 ; \mathrm{z}=2 \sin \mathrm{\pi t}$
Where x, y, z are measured in metre and t in second. The acceleration of the particle at t=3s is
(A)  $36\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}} {\mathrm{ms}}^{-2}$
(B)  $36\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\mathrm{\pi }\hat{\mathrm{k}} {\mathrm{ms}}^{-2}$
(C)  $36\hat{\mathrm{i}}+2\hat{\mathrm{j}} {\mathrm{ms}}^{-2}$
(D)  $12\hat{\mathrm{i}}+2\hat{\mathrm{j}} {\mathrm{ms}}^{-2}$

A particle moves in the \(xy\) plane and at time \(t\) is as the point whose coordinates are \(t^2, t^3-2t.\) Then at what instant of time will its velocity and acceleration vectors be perpendicular to each other?
1. \(\frac{1}{3}~\text{s}\)
2. \(\frac{2}{3}~\text{s}\)
3. \(\frac{3}{2}~\text{s}\)
4. Never

The coordinates of a moving particle at a time t, are given by, x=5 sin 10t, y=5 cos 10t. The speed of the particle is:
(1)  25
(2)  50
(3)  10
(4)  None

A particle moves in the x-y plane with velocity ${\mathrm{v}}_{\mathrm{x}}=8\mathrm{t}-2$ and ${\mathrm{v}}_{\mathrm{y}}=2$. If it passes through the point x=14 and y=4 at t=2 sec. The equation of the path is
(A)  $\mathrm{x}={\mathrm{y}}^2-\mathrm{y}+2$
(B)  $\mathrm{x}=\mathrm{y}+2$
(C)  $\mathrm{x}={\mathrm{y}}^2+2$
(D)  $\mathrm{x}={\mathrm{y}}^2+\mathrm{y}+2$

A motor boat of mass m moving along a lake with velocity ${\mathrm{V}}_0$. At t=0, the engine of the boat is shut down. Magnitude of resistance force offered to the boat is equal to rV. (V is instantaneous speed). What is the total distance covered till it stops completely? $\left[\mathrm{Hint}: \mathrm{F}\left(\mathrm{x}\right)=\mathrm{mV}\frac{\mathrm{dV}}{\mathrm{dx}}=-\mathrm{rV}\right]$
(A)  ${\mathrm{mV}}_0/\mathrm{r}$
(B)  $3 {\mathrm{mV}}_0/2\mathrm{r}$
(C)  ${\mathrm{mV}}_0/2\mathrm{r}$
(D)$2{\mathrm{mV}}_0/\mathrm{r}$

A particle is moving along positive x-axis. Its position varies as $\mathrm{x}={\mathrm{t}}^3-3{\mathrm{t}}^2+12\mathrm{t}+20$, where x is in meters and t is in seconds.
Initial velocity of the particle is
(1)  1 m/s
(2)  3 m/s
(3)  12 m/s
(4)  20 m/s

A particle is moving along positive x-axis. Its position varies as $\mathrm{x}={\mathrm{t}}^3-3{\mathrm{t}}^2+12\mathrm{t}+20$, where x is in meters and t is in seconds.
Initial acceleration of the particle is
(1)  Zero
(2)  $1 \mathrm{m}/{\mathrm{s}}^2$
(3)  $-3 \mathrm{m}/{\mathrm{s}}^2$
(4)  $-6 \mathrm{m}/{\mathrm{s}}^2$

A particle is moving along positive x-axis. Its position varies as $\mathrm{x}={\mathrm{t}}^3-3{\mathrm{t}}^2+12\mathrm{t}+20$, where x is in meters and t is in seconds.
Velocity of the particle when its acceleration zero is
(1)  1 m/s
(2)  3 m/s
(3)  6 m/s
(4)  9 m/s