Physics Motion in Straight Line

Physics Motion in Straight Line Motion in a Straight Line

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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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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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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Physics Motion in Straight Line Motion in a Straight Line

Contributor-Level 10

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

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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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2 months ago

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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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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2 months ago

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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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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2 months ago

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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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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2 months ago

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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 max^m height, = $\frac{1}{n} s e c$

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

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

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

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2 months ago

A particle is moving in a straight line such that its velocity is increasing at 5 ms^-1 per meter. The acceleration of the particle is _________ms^-2 at a point where its velocity is 20 ms^-1.

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Contributor-Level 10

$t a n θ = \frac{d v}{d x} = 5 (m / s) / m$

$\Rightarrow d v = 5 d x \Rightarrow \frac{d v}{d t} = 5 \frac{d x}{d t}$

$\Rightarrow a = 5 v = 5 * 20 = 100 m / s^{2}$

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2 months ago

A force on an object of mass 100g is $(10 \hat{i} + 5 \hat{j}) N .$ The position of that object at t = 2s is $(a \hat{i} + b \hat{j}) m$ after starting from rest. The value of $\frac{a}{b}$ will be

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