For a particle displacement time relation is \( t = \sqrt{x} + 3 \). Its displacement when its velocity is zero:
(1999)
1. 2 m
2. 4 m
4. None of these
View Answer
Concept: Velocity is the time derivative of displacement. First, express \( x \) as a function of \( t \): from \( t = \sqrt{x} + 3 \), we get \( \sqrt{x} = t - 3 \), so \( x = (t-3)^2 \). Then find velocity \( v = \frac{dx}{dt} = 2t-6 \). Set \( v=0 \) to find when it is at rest: \( 2t-6=0 \) implies \( t=3 \text{ s} \). Substitute \( t=3 \text{ s} \) back into the displacement equation: \( x(3) = (3-3)^2 = 0 \text{ m} \).
A bus travelling the first one-third distance at a speed of 10 km/h, the next one-third at 20 km/h and at last one-third at 60 km/h. The average speed of the bus is:
(1997)
1. 9 km/h
2. 16 km/h
3. 18 km/h
4. 48 km/h
View Answer
Concept: Average speed is total distance over total time. Let total distance be \( D \). The time taken for each one-third distance is \( t_1 = \frac{D/3}{10} = \frac{D}{30} \), \( t_2 = \frac{D/3}{20} = \frac{D}{60} \), \( t_3 = \frac{D/3}{60} = \frac{D}{180} \). Total time \( T = t_1+t_2+t_3 = D \left( \frac{6+3+1}{180} \right) = \frac{D}{18} \). Average speed \( = \frac{D}{T} = \frac{D}{D/18} = 18 \text{ km/h} \).
A car moves a distance of 200 m. It covers the first half of the distance at speed 40 km/h and the second half of distance at speed \( v \). The average speed is 48 km/h. The value of \( v \) is:
(1991)
1. 56 km/h
2. 60 km/h
3. 50 km/h
4. 48 km/h
View Answer
Concept: Average speed for two equal distance segments is \( v_{avg} = \frac{2v_1 v_2}{v_1 + v_2} \). Given \( v_{avg} = 48 \text{ km/h} \), \( v_1 = 40 \text{ km/h} \), and \( v_2 = v \). Substitute these values: \( 48 = \frac{2 \times 40 \times v}{40 + v} \). Simplify and solve for \( v \): \( 48(40+v) = 80v \) leads to \( 32v = 1920 \), so \( v = 60 \text{ km/h} \).
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 \(30\text{ km/h}\) and \(40\text{ km/h}\) respectively. The velocity of the car midway between P and Q is:
(1988)
1. 33.3 km/h
2. \(20\sqrt{2}\text{ km/h}\)
3. \(25\sqrt{2}\text{ Km/h}\)
4. 35 km/h
View Answer
Concept: Equations of motion under uniform acceleration.
Formula: \(v^2 = u^2 + 2as\).
Solution: Let \(u_P=30\), \(v_Q=40\) and distance be \(s\). \(v_Q^2 = u_P^2 + 2as\) gives \(40^2 = 30^2 + 2as\) => \(1600 = 900 + 2as\) => \(2as = 700\) => \(as = 350\). For the midway point, \(v_m^2 = u_P^2 + 2a(s/2) = u_P^2 + as = 30^2 + 350 = 900 + 350 = 1250\). So, \(v_m = sqrt{1250} = 25\sqrt{2}\text{ km/h}\).
A ball is thrown vertically downward with a velocity of \(20\text{ m/s}\) from the top of a tower. It hits the ground after some time with a velocity of \(80\text{ m/s}\). The height of the tower is : \((g = 10\text{ m/s}^2)\)
(2020)
1. 340 m
2. 320 m
3. 300 m
4. 360 m
View Answer
Concept: Equations of motion under gravity.
Formula: \(v^2 = u^2 + 2gh\).
Solution: Given \(u = 20\text{ m/s}\,\text{ }v = 80\text{ m/s}\,\text{ }g = 10\text{ m/s}^2\). Substituting these values: \(80^2 = 20^2 + 2(10)h\). \(6400 = 400 + 20h\). \(6000 = 20h\) => \(h = 300\text{ m}\).
A person sitting in the ground floor of a building notices through the window, of height \(1.5\text{ m}\), a ball dropped from the roof of the building crosses the window in \(0.1\text{ s}\). What is the velocity of the ball when it is at the topmost point of the window? \((g = 10\text{ m/s}^2)\)
(2020-Covid)
1. 14.5 m/s
2. 4.5 m/s
3. 20 m/s
4. 15.5 m/s
View Answer
Concept: Equations of motion under gravity for a specific interval.
Formula: \(h = ut + \frac{1}{2}gt^2\).
Solution: Let \(u\) be velocity at window top. Given \(h=1.5\text{ m}\,\text{ }t=0.1\text{ s}\,\text{ }g=10\text{ m/s}^2\). \(1.5 = u(0.1) + \frac{1}{2}(10)(0.1)^2\). \(1.5 = 0.1u + 0.05\). \(1.45 = 0.1u\) => \(u = 14.5\text{ m/s}\).
A stone falls freely under gravity. It covers distances \(h_1, h_2\) and \(h_3\) in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between \(h_1, h_2\) and \(h_3\) is:
(2013)
1. \(h_1 = h_2 = h_3\)
2. \(h_1 = 2h_2 = 3h_3\)
3. \(h_1 = \frac{h_2}{3} = \frac{h_3}{5}\)
4. \(h_2 = 3h_1 and h_3 = 3h_2 \)
View Answer
Concept: Distances covered by a freely falling body in equal successive time intervals.
Rule: For a body falling from rest, distances in successive equal time intervals are in ratio 1:3:5:...
Solution: \(h_1:h_2:h_3 = 1:3:5\). This implies \(h_2 = 3h_1\) and \(h_3 = 5h_1\). Therefore, \(h_1 = h_2/3 = h_3/5\).
A boy standing at the top of a tower of \(20\text{ m}\text{ height drops a stone. Assuming } g = 10\text{ m/s}^2\text{, the velocity with which it hits the ground is:}\)
[2011 Pre]
1. 10.0 m/s
2. 20.0 m/s
3. 40.0 m/s
4. 5.0 m/s
View Answer
Concept: Free fall under gravity.
Formula: \(v^2 = u^2 + 2gh\).
Solution: Given \(u=0\) (dropped), \(h=20\text{ m}\,\text{ }g=10\text{ m/s}^2\). \(v^2 = 0^2 + 2(10)(20) = 400\). So, \(v = \sqrt{400} = 20\text{ m/s}\).
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 \(g=10\text{ m/s}^2\), \(900 = 12v\) => \(v = 75\text{ m/s}\).