A particle moves a distance (x) in time (t) according to equation \(x = (t + 5)^{-1}\). The acceleration of particle is proportional to:
(2010 Pre)
1. \(\text{Velocity}^{2/3}\)
2. \(\text{Velocity}^{3/2}\)
3. \((\text{Distance})^2\)
4. \((\text{Distance})^{-2}\)
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Given \(x = (t + 5)^{-1}\). Velocity \(v = \frac{dx}{dt} = -(t + 5)^{-2}\). Acceleration \(a = \frac{dv}{dt} = 2(t + 5)^{-3}\). From \(v = -(t + 5)^{-2}\), we have \((t+5)^{-1} = ((-v)^{-1/2})\). So \(a = 2((t+5)^{-1})^3 = 2((-v)^{-1/2})^3 = 2(-v)^{3/2}\). Thus, \(a \propto (\text{Velocity})^{3/2}\).
A particle starts its motion from rest under the action of a constant force. If the distance covered in first \(10\) seconds is \(S_1\) and that covered in the first \(20\)seconds is \(S_2\), then:
(2009)
1. \(S_2 = 3S_1\)
2. \(S_2 = 4S_1\)
3. \(S_2 = S_1\)
4. \(S_2 = 2S_1\)
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For constant acceleration from rest, distance \(S = \frac{1}{2}at^2\). So \(S \propto t^2\). For \(t=10 \text{ s}\), \(S_1 = \frac{1}{2}a(10)^2 = 50a\). For \(t=20 \text{ s}\), \(S_2 = \frac{1}{2}a(20)^2 = 200a\). Therefore, \(S_2 = 4S_1\).
The distance travelled by a particle starting from rest and moving with an acceleration \(\frac{4}{3} \text{ m s}^{-2}\), in the third second is
(2008)
1. \(\frac{10}{3} \text{ m}\)
2. \(\frac{19}{3} \text{ m}\)
3. \(6 \text{ m}\)
4. \(4 \text{ m}\)
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The distance in the \(n^{\text{th}}\) second is given by \(S_n = u + \frac{a}{2}(2n - 1)\). Here (u=0), \(a = \frac{4}{3} \text{ m s}^{-2}\) and (n=3). So \(S_3 = 0 + \frac{4/3}{2}(2 times 3 - 1) = \frac{2}{3}(5) = \frac{10}{3} \text{ m}\).
A particle moves in a straight line with a constant acceleration. It changes its velocity from \(10 \text{ m s}^{-1}\) to \(20 \text{ m s}^{-1}\) while passing through a distance \(135 \text{ m}\) in (t) second. The value of (t) is
(2008)
View Answer
Given \(u = 10 \text{ m/s}\), \(v = 20 \text{ m/s}\), (\S = 135 \text{ m}\). Using \(v^2 = u^2 + 2aS\), \((20)^2 = (10)^2 + 2a(135) \Rightarrow 400 = 100 + 270a \Rightarrow a = \frac{300}{270} = \frac{10}{9} \text{ m/s}^2\). Now use \(v = u + at\), \(20 = 10 + \frac{10}{9}t \Rightarrow 10 = \frac{10}{9}t \Rightarrow t = 9 \text{ s}\).
A ball is thrown vertically upward. It has a speed of \(10\text{ m/s}\) when it has reached one half of its maximum height. How high does the ball rise? (Taking \(‘g’ = 10\text{ m/s}^2\)\)
(2005)
1. \(6\text{ m}\)
2. \(10\text{ m}\)
3. \(14\text{ m}\)
4. \(18/text{ m}\)
View Answer
Concept: Vertical motion under gravity. Let \(H\) be maximum height, \(u\) be initial velocity.
Formula: \(v^2 = u^2 - 2gh\). At \(H\), \(v=0 \Rightarrow u^2 = 2gH\).
At \(H/2\), \(10^2 = u^2 - 2g(H/2) = u^2 - gH\).
Substitute \(u^2 = 2gH\): \(100 = 2gH - gH = gH\).
Given \(g=10\text{ m/s}^2\), so \(100 = 10H \Rightarrow H = 10\text{ m}\).
A man throws ball with the same speed vertically upwards one after the other at an interval of \(2\text{ seconds}\)). What should be the speed of the throw so that more than two balls are in the sky at any time? (Given \(g = 9.8 m/s^2\)
(2003)
1. More than \(19.6\text{ m/s}\)
2. At least \(9.8\text{ m/s}\)
3. Any speed less than \(19.6\text{ m/s}\)
4. Only with speed \(19.6\text{ m/s}\)
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Concept: Time of flight for vertical motion. Let \(\Delta t = 2\text{ s}\)) be the throwing interval.
Formula: Time of flight \(T = 2u/g\).
For more than two balls to be in the air, the time of flight of each ball must be greater than twice the interval: \(T > 2\Delta t\).
So, \(2u/g > 2 \times 2 = 4\text{ s}\)).
\(u > 2g = 2 \times 9.8 = 19.6\text{ m/s}\).
A body dropped from a height \(h\) with initial velocity zero, strikes the ground with a velocity \(3\text{ m/s}\)). Another body of same mass dropped from the same height \(h\) with an initial velocity of \(4\text{ m/s}\)). The final velocity of second mass, with which it strikes the ground is:
(1996)
1. \(5\text{ m/s}\)
2. \(12\text{ m/s}\)
3. \(3\text{ m/s}\)
4. \(4\text{ m/s}\)
View Answer
Concept: Equations of motion under constant gravity.
Formula: \(v_f^2 = v_i^2 + 2gh\).
For the first body: \(v_i = 0\), \(v_f = 3\text{ m/s}\). So, \(3^2 = 0^2 + 2gh \Rightarrow 2gh = 9\).
For the second body: \(v_i = 4\text{ m/s}\). The final velocity is \(v_f'\).
\((v_f')^2 = 4^2 + 2gh = 16 + 9 = 25\).
Therefore, \(v_f' = \sqrt{25} = 5\text{ m/s}\).
The ‘x’ and ‘y’ coordinates of the particle at any time are \(x = 5t – 2t^2\) and \(y = 10t\), respectively, where ‘x’ and ‘y’ are in metres and ‘t’ in seconds. The acceleration of the particle at \(t = 2\text{ s}\) is:
(2017-Delhi)
1. \(5\text{ m/s}^2\)
2. \(-4\text{ m/s}^2\)
3. \(-8\text{ m/s}^2\)
4. \(0\)
View Answer
Given \(x = 5t - 2t^2\) and \(y = 10t\). Differentiating twice with respect to time to find acceleration. \(v_x = \frac{dx}{dt} = 5 - 4t\), \(a_x = \frac{dv_x}{dt} = -4\text{ m/s}^2\). \(v_y = \frac{dy}{dt} = 10\), \(a_y = \frac{dv_y}{dt} = 0\text{ m/s}^2\). The acceleration vector is \(\vec{a} = -4\hat{i}\). The x-component of acceleration is \(-4\text{ m/s}^2\), which is constant.