The displacement of a particle is given by $\mathrm{y}=\mathrm{a}+\mathrm{bt}+{\mathrm{ct}}^2-{\mathrm{dt}}^4$. The initial velocity and initial acceleration, respectively, are: (\(Given: v=\frac{dx}{dt}~and~a=\frac{d^2x}{dt^2}\))

1.  b, -4d

2.  -d, 2c

3.  b, 2c

4.  2c, -4d

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 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

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)  $0$

Temperature of a body varies with time as $\mathrm{T}=\left({\mathrm{T}}_0+{\mathrm{at}}^2+\mathrm{b} \mathrm{sint}\right)\mathrm{K}$, where ${\mathrm{T}}_0$ is the temperature in Kelvin at $\mathrm{t}=0 \sec \& \mathrm{a}=2/\mathrm{\pi } \mathrm{K}/{\mathrm{s}}^2 \& \mathrm{b}=-\mathrm{4 }\mathrm{K}$, then the rate of change of temperature $\left(\frac{\mathrm{dT}}{\mathrm{dt}}\right)$ at $\mathrm{t}=\mathrm{\pi } \sec$ is:

1.  $8 \mathrm{K}$

2.  ${80}^{} \mathrm{K}$

3.  $8 \mathrm{K}/\sec$

4.  ${80}^{} \mathrm{K}/\sec$

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)  $\frac{\mathrm{s}}{{\mathrm{t}}^2}$

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 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

The instantaneous velocity (defined as $v=\frac{ds}{dt}$) at time $t=\frac{\mathrm{\pi }}{2}$ of a particle, whose position equation is given as  s(t)=12 tan$\left(\frac{t}{2}+\mathrm{\pi }\right)$m, is

1. 12 m/s
2. 12$\sqrt{2}$ m/s
3. 6 m/s
4. \(6\sqrt2\) m/s