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

If \(f \left(x\right) = x^{2} - 2 x + 4\), then \(f(x)\) has:

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. \(8~\text{K}\)
2. \(80~\text{K}\)
3. \(8~\text{K/sec}\)
4. \(80~\text{K/sec}\)

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 \(x\text-\)axis. Its position varies as \(x = t^3-3t^2+12t+20,\) where \(x\) is in meters and \(t\) is in seconds. The velocity of the particle when its acceleration zero is:
1. \(1~\text{m/s}\)
2. \(3~\text{m/s}\)
3. \(6~\text{m/s}\)
4. \(9~\text{m/s}\)

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