A car is moving along a straight road with a uniform acceleration. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. The velocity of the car midway between P and Q is :
1. 33.3 km/hr
2. \(20\sqrt{3} km / hr\)
3. /(25\sqrt{2} km / hr/)
4. 35 km/hr
View Answer
For uniform acceleration, the velocity midway between two points is given by the formula: \(v_{mid} = \sqrt{\frac{v_1^2 + v_2^2}{2}} = \sqrt{\frac{30^2 + 40^2}{2}} = \sqrt{\frac{2500}{2}} = 25\sqrt{2}\text{ km/hr}\).
A paratrooper jumps from a height \(H\). The parachute can provide a uniform deceleration of \(2\text{ ms}^{-2}\). The height above the ground at which the parachute should be opened so that he touches ground with zero speed is (take \(g = 10\text{ ms}^{-2}\)):
1. \(\frac{H}{6}\)
2. \(\frac{4H}{5}\)
3. \(\frac{5H}{6}\)
4. \(\frac{6H}{7}\)
View Answer
Let \(h\) be free fall and \(y\) be decelerating height, so \(H = h + y\). Speed before parachute opens: \(v^2 = 2gh = 20h\). Deceleration phase: \(0 = v^2 - 2ay = v^2 - 4y\), which gives \(20h = 4y \Rightarrow y = 5h\). Since \(H = 6h\), we find \(y = \frac{5H}{6}\).
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 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.
A particle has initial velocity \(3\hat{i} + 4\hat{j}\) and has acceleration \(0.4\hat{i} + 0.3\hat{j}\). Its speed after 10 s is:
(2010 Pre)
1. \(10\text{ units}\)
2. \(7\text{ units}\)
3. \(7\sqrt{2}\text{ units}\)
4. \(8.5\text{ units}\)
View Answer
Initial velocity \(vec{v}_0 = 3\hat{i} + 4\hat{j}\). Acceleration \(\vec{a} = 0.4\hat{i} + 0.3\hat{j}\). Time \(t = 10\text{ s}\). Using \(\vec{v} = \vec{v}_0 + \vec{a}t\), we get \(\vec{v} = (3\hat{i} + 4\hat{j}) + (0.4\hat{i} + 0.3\hat{j})(10) = (3\hat{i} + 4\hat{j}) + (4\hat{i} + 3\hat{j}) = 7\hat{i} + 7\hat{j}\). Speed is the magnitude of velocity: \(|\vec{v}| = \sqrt{7^2 + 7^2} = \sqrt{49+49} = \sqrt{98} = 7\sqrt{2}\).
Two boys are standing at the ends A and B of a ground, where \(AB = a\). The boy at B starts running in a direction perpendicular to AB with velocity \(v_1\). The boy at A starts running simultaneously with velocity \(v\) and catches the other boy in a time t, where t is:
(2005)
1. \(\frac{a}{\sqrt{v^2+v_1^2}}\)
2. \(\frac{a}{\sqrt{v^2-v_1^2}}\)
3. \(a/(v-v_1)\)
4. \(a/(v+v_1)\)
View Answer
Let B be at \((0,0)\) and A at \((a,0)\) at \(t=0\). Boy B's position at time \(t\) is \(\vec{r}_B = v_1 t \hat{j}\). Boy A moves with velocity \(\vec{v}_A = v_{Ax}\hat{i} + v_{Ay}\hat{j}\). For A to catch B, their positions must be equal at time \(t\). So, \(a\hat{i} + \vec{v}_A t = v_1 t \hat{j}\). This implies \(v_{Ax} = -a/t\) and \(v_{Ay} = v_1\). The magnitude of A's velocity is \(v = |\vec{v}_A| = \sqrt{v_{Ax}^2 + v_{Ay}^2}\). So, \(v^2 = (-a/t)^2 + v_1^2\). Rearranging for \(t\): \(t^2 = \frac{a^2}{v^2 - v_1^2}\), hence \(t = \frac{a}{\sqrt{v^2 - v_1^2}}\).