A ball is dropped from a high rise platform at \(t = 0\) starting from rest. After 6 seconds another ball is thrown downwards from the same platform with a speed \(v\). The two balls meet at \(t = 18\text{s}\). What is the value of \(v\)?
[2010 Pre]
1. 75 m/s
2. 55 m/s
3. 40 m/s
4. 60 m/s
View Answer
Concept: Motion under gravity and meeting condition.
Formula: \(h = ut + \frac{1}{2}gt^2\).
Solution: For ball 1 (dropped at \(t=0\)): \(h_1 = \frac{1}{2}g(18)^2 = 162g\). For ball 2 (thrown at \(t=6\text{ s}\), travels for \(12\text{ s}\)): \(h_2 = v(12) + \frac{1}{2}g(12)^2 = 12v + 72g\). When they meet, \(h_1 = h_2\): \(162g = 12v + 72g\). \(90g = 12v\). Using \(g=10\text{ m/s}^2\), \(900 = 12v\) => \(v = 75\text{ m/s}\).
The velocity of train increases uniformly from \(20 \text{ km/h}\) to \(60 \text{ km/h}\) in 4 hours. The distance travelled by the train during this period, is:
(1994)
1. 160 Km
2. 180 Km
3. 100 Km
4. 120 Km
View Answer
Given initial velocity \(u = 20 \text{ km/h}\), final velocity \(v = 60 \text{ km/h}\), and time \(t = 4 \text{ h}\). For uniform acceleration, the distance \(S = \frac{u+v}{2}t\). Plugging in the values, \(S = \frac{20 + 60}{2} \times 4 = \frac{80}{2} \times 4 = 40 \times 4 = 160 \text{ km}\).
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. \(S_2 = 2S_1\)
View Answer
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. \(4 \text{ m}\)
View Answer
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 \(v = u + at\), \(20 = 10 + \frac{10}{9}t \Rightarrow 10 = \frac{10}{9}t \Rightarrow t = 9 \text{ s}\).
If a car at rest accelerates uniformly to a speed of \(144 \text{ km/h}\) in \(20 \text{ sec}\), it covers a distance of:
(1997)
1. \(1440 \text{ cm}\)
2. \(2980 \text{ cm}\)
3. \(20 \text{ m}\)
4. \(400 \text{ m}\)
View Answer
Given (u=0), \(v = 144 \text{ km/h} = 144 \times \frac{5}{18} = 40 \text{ m/s}\), \(t = 20 \text{ s}\). Using \(S = \frac{u+v}{2}t), we get \(S = \frac{0+40}{2} \times 20 = 20 \times 20 = 400 \text{ m}\).
The acceleration of a particle is increasing linearly with time (t) as \(bt\). The particle starts from origin with an initial velocity \(v_0\). The distance travelled by the particle in time (t) will be:
(1995)
1. \(v_0 t + \frac{bt^2}{3}\)
2. \(v_0 t + \frac{bt^2}{2}\)
3. \(v_0 t + \frac{bt^3}{6}\)
4. \(v_0 t + \frac{bt^3}{3}\)
View Answer
Given \(a = \frac{dv}{dt} = bt\). Integrating, \(v = \int bt , dt = \frac{1}{2}bt^2 + C_1\). Since \(v=v_0\) at \(t=0\), \(C_1 = v_0\). So \(v = v_0 + \frac{1}{2}bt^2\). Given \(v = \frac{dx}{dt}\). Integrating again, \(x = \int (v_0 + \frac{1}{2}bt^2\) , \(dt = v_0 t + \frac{1}{2}b \frac{t^3}{3} + C_2\). Since (x=0) at (t=0), (C_2 = 0). Thus, \(x = v_0 t + \frac{bt^3}{6}\).
The position vector of a particle \(\vec{R}\) as a function of time is given by: \(\vec{R} = 4sin(2\pi t)\hat{i} + 4cos(2\pi t)\hat{j}\) Where R is in metres, t is in seconds and \(\hat{i}\) and \(\hat{j}\) denote unit vectors along x and y-direction, respectively. Which one of the following statements is wrong for the motion of particle?
(2015)
1. Path of the particle is a circle of radius 4 metre
2. Acceleration vectors is along \(-vec{R}\)
3. Magnitude of acceleration vector is \(\frac{V^2}{R}\) where V is the velocity of particle.
4. Magnitude of the velocity of particle is 8 metre/second
View Answer
From \(\vec{R} = 4sin(2\pi t)\hat{i} + 4cos(2\pi t)\hat{j}\), \(x=4sin(2\pi t)\) and \(y=4cos(2\pi t)\). \(x^2+y^2=16\) implies a circle of radius 4m. \(\vec{V} = 8\pi cos(2\pi t)\hat{i} - 8\pi sin(2\pi t)\hat{j}\). \(|\vec{V}| = 8\pi\text{ m/s}\). \(\vec{a} = -16\pi^2sin(2\pi t)\hat{i} - 16\pi^2cos(2\pi t)\hat{j} = -4\pi^2 \vec{R}\). So \(\vec{a}\) is along \(-\vec{R}\). Also, \(|\vec{a}| = 16\pi^2\) and \(\frac{V^2}{R} = \frac{(8\pi)^2}{4} = 16\pi^2\). Therefore, (a), (b), (c) are correct. (d) is wrong because \(|\vec{V}| = 8\pi\text{ m/s}\), not 8 m/s.