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Physics Motion in Straight Line

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4 months ago

Physics Motion in Straight Line Class 11th

A particle is moving with constant acceleration 'a'. Following graph shows $ν^{2}$ versus x (displacement) plot. The acceleration of the particle is______ m/s².

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Vishal Baghel

Contributor-Level 10

Equation from the given graph:

$\begin{array}{l} v^{2} = 2 x + 20 \\ v^{2} = 2 x + 20 \\ \Rightarrow 2 V \frac{d V}{d t} = 2 V \end{array}$

=>a = 1m/s²

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New answer posted

4 months ago

Physics Motion in Straight Line Class 11th

Water drops are falling from a nozzle of a shower onto the floor, from a height of 9.8m. The drops fall at a regular interval of time. When the first drop strikes the floor, at that instant, the third drop begins to fall. Locate the position of second drop from the floor when the first drop strikes the floor.

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alok kumar singh

Contributor-Level 10

Time taken by first drop to reach the ground,

$t_{1} = \sqrt[]{\frac{2 h}{g}} = \sqrt[]{\frac{2 * 9.8}{9.8}} = \sqrt[]{2} s$

Time for which second drop has fallen,

$t_{2} = t_{1} - \frac{t_{1}}{2} = \sqrt[]{2} - \frac{1}{\sqrt[]{2}} s$

Height of the second drop

$= 9.8 - \frac{1}{2} * 9.8 * {(\sqrt[]{2} - \frac{1}{\sqrt[]{2}})}^{2} = 7.35 m$

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New answer posted

4 months ago

Physics Motion in Straight Line Class 11th

If the velocity of a body related to displacement x is given by $υ = \sqrt[]{5000 + 24 x} m / s,$ then the acceleration of the body is…….

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Vishal Baghel

Contributor-Level 10

$a = V \frac{d v}{d x}$

$\Rightarrow a = \sqrt[]{5000 + 24 x} * \frac{12}{\sqrt[]{5000 + 24 x}}$

=> a = 12 m/s²

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4 months ago

Physics Motion in Straight Line Class 11th

Two buses P AN Q start form a point at the same time and move in a straight line and their positions are represented by X_p (t) = at + bt² and X_Q (t) = ft – t². At what time, both the buses have same velocity?

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Raj Pandey

Contributor-Level 9

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ - (i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ - (ii)
(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

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5 months ago

Physics Motion in Straight Line Motion in a Straight Line

Two spherical balls having equal masses with radius of 5cm each are thrown upwards along the same vertical direction at an interval of 3s with the same initial velocity of 35m/s, then these balls collide at a height of…………. m. (Take g = 10 m/s²)

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Vishal Baghel

Contributor-Level 10

y₁ = y₂

=> $35 t - \frac{1}{2} g t^{2} = 35 (t - 3) - \frac{1}{2} g {(t - 3)}^{2}$

=> 35 * 3 = $\frac{1}{2} g (t + t - 3) * 3$

105 = $\frac{10}{2} * 3 * (2 t - 3)$

t = 5

$y_{1} = 35 * 5 - \frac{1}{2} * 10 * 5^{2}$

= 50 m

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5 months ago

Physics Motion in Straight Line Class 11th

A bullet is shot vertically downwards with an initial velocity of 100m/s a certain height. Within 10s, the bullet reaches the ground and instantaneously comes to rest due to the perfectly inelastic collision. The velocity –time curve for total time t = 20s will be:

(Take g = 10m/s²).

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alok kumar singh

Contributor-Level 10

V = u + at -> v = -100 - 10 * 10 = -200 m/s

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New answer posted

5 months ago

Physics Motion in Straight Line Motion in a Straight Line

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alok kumar singh

Contributor-Level 10

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ -(i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ -(ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$ &

...more

$X_{P} (t) = α t + β t^{2}$

$V_{P} (t) = α + 2 β t$ -(i)

$X_{Q} (t) = f t - t^{2}$

$V_{Q} (t) = f - 2 t$ -(ii)

(i) = (ii)

$α + 2 β t = f - 2 t \Rightarrow t (2 β + 2) = f - α$

$t = \frac{f - α}{2 (β + 1)}$

less

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New answer posted

5 months ago

Physics Motion in Straight Line Class 11th

A ball is released from a height h. If t₁ and t₂ be the time required to complete first half and second half of the distance respectively. Then, choose the correct relation between t₁ and t₂.

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Payal Gupta

Contributor-Level 10

$\frac{h}{2} = \frac{1}{2} g t_{1}^{2}$

h = $\frac{1}{2} g {(t_{1} + t_{2})}^{2}$

from (i) and (ii) :-

$\frac{1}{2} g t_{1}^{2} = \frac{g}{4} {(t_{1} + t_{2})}^{2}$

$2 t_{1}^{2} = {(t_{1} + t_{2})}^{2}$

$\sqrt[]{2} t_{1} = t_{1} + t_{2}$

$t_{1} = \frac{t_{2}}{\sqrt[]{2} - 1} = (\sqrt[]{2} + 1) t_{2}$

$t_{2} = (\sqrt[]{2} - 1) t_{1}$

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New answer posted

5 months ago

Physics Motion in Straight Line Class 11th

A juggler throws ball vertically upwards with same initial velocity in air. When the first ball reaches its highest positions he throws the next ball. Assuming the juggler throws n balls per second, the maximum height the balls can reach is

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Payal Gupta

Contributor-Level 10

Time to reach at max height t = $\frac{u}{g} n o .$ of balls thrown in 1 sec = n.

So, time taken by each ball to reach maxm height, = $\frac{1}{n} s e c$

i.e. $\frac{u}{g} = \frac{1}{n} \Rightarrow u = \frac{g}{n}$ So, hmax = $\frac{u^{2}}{2 g}$

= $\frac{g^{2}}{2 g n^{2}}$

$= \frac{g}{2 n^{2}}$

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New answer posted

5 months ago

Physics Motion in Straight Line Class 11th

Sand is being dropped from a stationary dropper at a rate of 0.5 kgs^-1 on a conveyor belt moving with a velocity of 5ms^-1. The power needed to keep the belt moving with the same velocity will be:

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alok kumar singh

Contributor-Level 10

$P = \vec{F} . \vec{v} = v \frac{d P}{d t} = v^{2} (\frac{d m}{d t}) = 0.5 * 25 = 12.5 W$

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