A bus starts from rest moving with an acceleration of \(2\text{ m/s}^2\). A cyclist, 96 m behind the bus starts simultaneously towards the bus at a speed of \(20\text{ m/s}\). After what time will bus overtake the cycle :
1. 8 s
2. 10 s
3. 12 s
4. 1 s
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Let the bus's position be \(x_b = (1/2)at^2 = t^2\). The cyclist's position (starting 96m behind) is \(x_c = 20t - 96\). For meeting, \(t^2 = 20t - 96 \Rightarrow t^2 - 20t + 96 = 0\). Factoring gives \((t-8)(t-12) = 0\), so \(t=8\)s or \(t=12\)s. At \(t=8\)s, cyclist is faster than bus. At \(t=12\)s, bus is faster, so it overtakes.
Two bodies are held separated by \(9.8\text{ m}\) vertically one above the other. They are released simultaneously to fall freely under gravity. After \(2\text{ s}\) the relative distance between them is :
1. 4.9 m
2. 19.6 m
3. 9.8 m
4. 39.2m
View Answer
Both bodies are released simultaneously and fall under gravity. Their acceleration is identical (\(g\)). Since their initial relative velocity is zero and relative acceleration is zero, their relative distance remains constant. Thus, after \(2\) s, the relative distance is still \(9.8\) m.
While sitting on a tree branch \(20\text{m}\) above the ground, you drop a walnut. When the walnut has fallen \(5\text{m}\) you throws a second walnut straight down. What initial speed must you give the second walnut if they are both to reach the ground at the same time? (g=\(10\text{m/s}^2\))
1. 5 m\(s^{-1}\)
2. 10 m\(s^{-1}\)
3. 15 m\(s^{-1}\)
4. None of these
View Answer
First walnut (W1): Time to fall 5m from rest is \(t_1 = \sqrt{2s/g} = \sqrt{2 \cdot 5/10} = 1\) s. Total time for W1 to reach ground from 20m is \(t_{total} = \sqrt{2 \cdot 20/10} = 2\) s. Second walnut (W2) must fall 15m in \(t_{W2} = t_{total} - t_1 = 2-1 = 1\) s. Using \(s = v_0t + (1/2)gt^2\): \(15 = v_0(1) + (1/2)(10)(1)^2 \Rightarrow 15 = v_0 + 5 \Rightarrow v_0 = 10\text{ m/s}\).
When a motorcycle moving with a uniform speed \(11\text{ m/s}\) is at a distance \(24\text{ m}\) from a car, the car starts from rest and moves with a uniform acceleration \(2\text{ m/s}^2\) away from the motorcycle. If the car begins motion at \(t = 0\), time at which the motorcycle will overtake the car is \(t = \):
1. \(8\text{ sec}\)
2. \(6\text{ sec}\)
3. \(3\text{ sec}\)
4. \(1.5\text{ sec}\)
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Distance equation for meeting: \(11t = 24 + \frac{1}{2}(2)t^2 \Rightarrow t^2 - 11t + 24 = 0\). Solving this quadratic equation gives \(t = 3\text{ s}\) and \(t = 8\text{ s}\). The first overtake occurs at \(t = 3\text{ s}\).
Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time \( t_1 \). On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time \( t_2 \). The time taken by her to walk up on the moving escalator will be:
(2017-Delhi)
1. \( \frac{t_1 t_2}{t_2 - t_1} \)
2. \( \frac{t_1 t_2}{t_2 + t_1} \)
3. \( t_2 - t_1 \)
4. \( \frac{t_1 + t_2}{2} \)
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Concept: Relative velocity.
If Preeti's speed is \( v_p \) and escalator's speed is \( v_e \), for total length \( L \), \( v_p = L/t_1 \) and \( v_e = L/t_2 \). When Preeti walks on moving escalator, effective speed is \( v_p + v_e \). Time taken \( T = L / (v_p + v_e) = L / (L/t_1 + L/t_2) = \frac{t_1 t_2}{t_1 + t_2} \).
Two cars P and Q start from a point at the same time in a straight line and their positions are represented by \( X_P(t) = at + bt^2 \) and \( X_Q(t) = ft – t^2 \). At what time do the cars have the same velocity?
(2016 – II)
1. \( \frac{a+f}{2(1+b)} \)
2. \( \frac{f-a}{2(1+b)} \)
3. \( \frac{a-f}{1+b} \)
4. \( \frac{a+f}{2(b-1)} \)
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Concept: Velocity is the time derivative of position. Calculate \( V_P(t) = \frac{dX_P}{dt} = a + 2bt \) and \( V_Q(t) = \frac{dX_Q}{dt} = f - 2t \). Equate \( V_P(t) = V_Q(t) \) to find time \( t \). \( a + 2bt = f - 2t \) ⇒ \( 2t(b+1) = f-a \), so \( t = \frac{f-a}{2(b+1)} \).