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The speed of a body moving with uniform acceleration is u. This speed is doubled while covering a distance S. When it covers an additional distance S, its speed would become

$\sqrt{3} u$

$\sqrt{5} u$

$\sqrt{11} u$

$\sqrt{7} u$

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Explanation

As $v^{2} = u^{2} + 2 a s$ ⇒ ${(2 u)}^{2} = u^{2} + 2 a s \Rightarrow 2 a s = 3 u^{2}$

Now, after covering an additional distance s, if velocity becomes v, then,

$v^{2} = u^{2} + 2 a (2 s) = u^{2} + 4 a s = u^{2} + 6 u^{2} = 7 u^{2}$

∴ $v = \sqrt{7} u$ .

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Two trains one of length 100 m and another of length 125 m, are moving in mutually opposite directions along parallel lines, meet each other, each with speed 10 m/s. If their acceleration are 0.3 m/s² and 0.2 m/s² respectively, then the time they take to pass each other will be

5 s

10 s

15 s

20 s

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Explanation

Relative velocity of one train w.r.t. other

= 10 + 10 = 20 m/s.

Relative acceleration =0.3+0.2=0.5 m/s²

If trains cross each other then from $s = u t + \frac{1}{2} a t^{2}$

$A s, s = s_{1} + s_{2} = 100 + 125 = 225$

⇒ $225 = 20 t + \frac{1}{2} \times 0 .5 \times t^{2}$ ⇒ $0 .5 t^{2} + 40 t - 450 = 0$

⇒ $t = - \frac{40 \pm \sqrt{1600 + 4 . (005) \times 450}}{1} = - 40 \pm 50$

∴ t = 10 sec (Taking +ve value).

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A body starts from rest with uniform acceleration. If its velocity after n second is v, then its displacement in the last two seconds is

$\frac{2 v (n + 1)}{n}$

$\frac{v (n + 1)}{n}$

$\frac{v (n - 1)}{n}$

$\frac{2 v (n - 1)}{n}$

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Explanation

$∵ v = 0 + n a \Rightarrow a = v / n$

Now, distance travelled in n sec. ⇒ $S_{n} = \frac{1}{2} a n^{2}$ and distance travelled in (n – 2) sec ⇒ $S_{n - 2} = \frac{1}{2} a {(n - 2)}^{2}$

∴ Distance travelled in last two seconds,

$= S_{n} - S_{n - 2} = \frac{1}{2} a n^{2} - \frac{1}{2} a {(n - 2)}^{2}$

$= \frac{a}{2} [n^{2} - {(n - 2)}^{2}] = \frac{a}{2} [n + (n - 2)] [n - (n - 2)]$

= $a (2 n - 2) = \frac{v}{n} (2 n - 2) = \frac{2 v (n - 1)}{n}$

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A point starts moving in a straight line with a certain acceleration. At a time t after beginning of motion the acceleration suddenly becomes retardation of the same value. The time in which the point returns to the initial point is

$\sqrt{2 t}$

$(2 + \sqrt{2}) t$

$\frac{t}{\sqrt{2}}$

Cannot be predicted unless acceleration is given

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Explanation

If a particle starts moving with an acceleration and then experiences an equal retardation, the time taken to return to the initial point is (2 + âˆš2) times the time taken to reach the highest point. This can be derived using kinematic equations and the symmetry of the motion.

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A bird flies for 4 s with a velocity of $| t - 2 | m / s$ in a straight line, where t is time in seconds. It covers a distance of

2 m

4 m

6 m

8 m

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A particle is projected with velocity $v_{0}$ along x-axis. The deceleration of the particle is proportional to the square of the distance from the origin i.e., $a = α x^{2} .$ The distance at which the particle stops is :

$\sqrt{\frac{3 v_{0}}{2 α}}$

${(\frac{3 v_{o}}{2 α})}^{\frac{1}{3}}$

$\sqrt{\frac{3 v {}_{0}^{2}}{2 α}}$

${(\frac{3 v {}_{0}^{2}}{2 α})}^{\frac{1}{3}}$

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Explanation

$a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x} = - α x^{2}$ (given)

⇒ $\int_{v_{0}}^{0} v d v = - α \int_{0}^{S} x^{2} d x$ ⇒ ${[\frac{v^{2}}{2}]}_{v_{0}}^{0} = - α {[\frac{x^{3}}{3}]}_{0}^{S}$

⇒ $\frac{v_{0}^{2}}{2} = \frac{α S^{3}}{3}$ ⇒ $S = {[\frac{3 v_{0}^{2}}{2 α}]}^{\frac{1}{3}}$

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A body is projected vertically up with a velocity v and after some time it returns to the point from which it was projected. The average velocity and average speed of the body for the total time of flight are

$\vec{v} / 2$ and v/2

0 and v/2

0 and 0

$\vec{v} / 2$ and 0

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Explanation

Average velocity = 0 because net displacement of the body is zero.

Average speed $= \frac{Total distance covered}{Time of flight} = \frac{2 H_{\max}}{2 u / g}$

⇒ $v_{a v} = \frac{2 u^{2} / 2 g}{2 u / g}$ ⇒ $v_{a v} = u / 2$

Velocity of projection = v (given)

∴ $v_{av} = v / 2$

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A stone is dropped from a height h. Simultaneously, another stone is thrown up from the ground which reaches a height 4 h. The two stones cross each other after time

$\sqrt{\frac{h}{8 g}}$

$\sqrt{8 g h}$

$\sqrt{2 g h}$

$\sqrt{\frac{h}{2 g}}$

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Four marbles are dropped from the top of a tower one after the other with an interval of one second. The first one reaches the ground after 4 seconds. When the first one reaches the ground the distances between the first and second, the second and third and the third and forth will be respectively

35, 25 and 15 m

30, 20 and 10 m

20, 10 and 5 m

40, 30 and 20 m

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A balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is

$4 \sqrt{\frac{h}{g}}$

$2 \sqrt{\frac{h}{g}}$

$\sqrt{\frac{2 h}{g}}$

$\sqrt{\frac{g}{h}}$

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Explanation

The velocity of balloon at height h, $v = \sqrt{2 (\frac{g}{8}) h}$

When the stone released from this balloon, it will go upward with velocity v = $\frac{\sqrt{g h}}{2}$ (Same as that of balloon). In this condition time taken by stone to reach the ground using 2nd equation of motion:

$- h = \frac{\sqrt{g h}}{2} t - \frac{1}{2} g t^{2} t^{2} - \sqrt{\frac{h}{g}} t - \frac{2 h}{g} = 0 t = \frac{\sqrt{\frac{h}{g}} \pm \sqrt{\frac{h}{g} + 4 \times \frac{2 h}{g}}}{2} t = \frac{\sqrt{\frac{h}{g}} \pm 3 \sqrt{\frac{h}{g}}}{2} t = \frac{\sqrt{\frac{h}{g}} + 3 \sqrt{\frac{h}{g}}}{2} = 2 \sqrt{\frac{h}{g}}$

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