Questions in differentiation

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The maximum height is reached in 5 seconds by a stone thrown vertically upwards and moving under the equation $10s = 10ut – 49{t}^{2}$, where s is in metre and t is in second. The value of u is
If $2t={{v}^{2}},$ then $\frac{dv}{dt}$ is equal to
If $t=\frac{{{v}^{2}}}{2}$ ,then $\left( -\frac{df}{dt} \right)$ is equal to, (where $f$ is acceleration)
The equation of motion of a particle moving along a straight line is $s=2$ ${{t}^{3}}-9{{t}^{2}}+12t$ , where the units of s and t are cm and sec. The acceleration of the particle will be zero after
A particle is moving in a straight line according to the formula $s={{t}^{2}}+8t+12.$ If s be measured in metre and t be measured in second, then the average velocity of the particle in third second is
A 10cm long rod AB moves with its ends on two mutually perpendicular straight lines OX and OY. If the end A be moving at the rate of $2cm/\sec $ , then when the distance of A from O is $8cm$ , the rate at which the end B is moving, is
If the radius of a circle increases from 3 cm to 3.2 cm, then the increase in the area of the circle is
If $y={{x}^{3}}+5$ and $x$ changes from 3 to 2.99, then the approximate change in y is
A particle is moving on a straight line, where its position s (in metre) is a function of time t (in seconds) given by $s=a{{t}^{2}}+bt+6,t\ge 0$ . If it is known that the particle comes to rest after 4 seconds at a distance of 16 metre from the starting position $(t=0)$ , then the retardation in its motion is
A particle is moving in a straight line according as $s=45\,t+11{{t}^{2}}-{{t}^{3}}$ then the time when it will come to rest, is

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