Questions in differentiation

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If the law of motion in a straight line is $s=\frac{1}{2}v\,t,$ then acceleration is
A point moves in a straight line during the time $t=0$ to $t=3$ according to the law $s=15t-2{{t}^{2}}$ . The average velocity is
The distance in seconds, described by a particle in t seconds is given by $s=a{{e}^{t}}+\frac{b}{{{e}^{t}}}$ . Then acceleration of the particle at time t is
The equation of motion of a stone, thrown vertically upwards is $s=ut-6.3{{t}^{2}},$ where the units of s and t are cm and sec. If the stone reaches at maximum height in 3 sec, then u =
A particle moves in a straight line so that its velocity at any point is given by ${{v}^{2}}=a+bx$ , where $a,b\ne 0$ are constants. The acceleration is
The volume V and depth x of water in a vessel are connected by the relation $V=5x-\frac{{{x}^{2}}}{6}$ and the volume of water is increasing at the rate of $5c{{m}^{3}}/\sec $ , when $x=2cm$ . The rate at which the depth of water is increasing, is
The equation of motion of a stone thrown vertically upward from the surface of a planet is given by $s=10\,\,t-3{{t}^{2}}$ , and the units of s and t are cm and sec respectively. The stone will return to the surface of the planet after
A body moves according to the formula $v=1+{{t}^{2}}$ , where v is the velocity at time t. The acceleration after 3 sec will be (v in cm/sec)
The length of the side of a square sheet of metal is increasing at the rate of $4cm/\sec $ . The rate at which the area of the sheet is increasing when the length of its side is 2 cm, is
The equations of motion of two stones thrown vertically upwards simultaneously are $s=19.6\,t-4.9\,{{t}^{2}}$ and $s=9.8\,t-4.9\,{{t}^{2}}$ respectively and the maximum height attained by the first one is h. When the height of the first stone is maximum, the height of the second stone will be

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