A car is travelling at \(72\text{ kmh}^{-1}\) and is \(20\text{ m}\) from a barrier when the driver puts on the brakes. The car hits the barrier 2s later. What is the magnitude of the constant deceleration?
Initial velocity \(u = 72\text{ km/h} = 20\text{ m/s}\). Using \(s = ut - \frac{1}{2}at^2\): \(20 = 20(2) - \frac{1}{2}a(2)^2 ⇒ 20 = 40 - 2a ⇒a = 10\text{ ms}^{-2}\).
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 :
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.
A boat takes 2 hours to go 10 km and come back in still water lake. The time taken for going 10 km upstream and coming back with water velocity of \(5\text{ km/h}\).
In still water, total distance is 20 km in 2 hours, so boat speed \(v_b = 10\text{ km/h}\). Water velocity \(v_w = 5\text{ km/h}\). Upstream speed \(v_b - v_w = 5\text{ km/h}\). Time upstream = \(10/5 = 2\) hours. Downstream speed \(v_b + v_w = 15\text{ km/h}\). Time downstream = \(10/15 = 2/3\) hours. Total time = \(2 + 2/3 = 8/3\) hours = \(160\) minutes.
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 :
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=\(10\text{m/s}^2\))
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 \(s = v_0t + (1/2)gt^2\): \(15 = v_0(1) + (1/2)(10)(1)^2 \Rightarrow 15 = v_0 + 5 \Rightarrow v_0 = 10\text{ m/s}\).
A particle is projected from horizontal ground with speed \(50\text{ ms}^{-1}\) at \(53^\circ\) with horizontal. Find time after which velocity of particle will be at \(45^\circ\) with horizontal for the second time.
Horizontal speed is \(u_x = u\cos(53^\circ) = 30\text{ m/s}\), and vertical speed is \(u_y = u\sin(53^\circ) = 40\text{ m/s}\). At \(45^\circ\) on the way down, \(v_y = -v_x = -30\text{ m/s}\). Using \(v_y = u_y - gt\), we get \(-30 = 40 - 10t\), which yields \(t = 7\text{ s}\).
A point moves linearly such that \(\frac{dv}{dt} = -\alpha v^{1/2}\), where \(\alpha\) is a positive constant. At start, \(v = v_0\). Find time taken by particle to stop?
Integrating \(v^{-1/2} dv = -\alpha dt\) from \(v_0\) to \(0\) gives \(2\sqrt{v_0} = \alpha t\), so the time taken to stop is \(t = \frac{2\sqrt{v_0}}{\alpha}\).
Two bodies separated by a distance of \(‘s’\) start moving towards each other with speeds of \(v\) and \(2v\) respectively. The uniform acceleration with which the first body should move so that they meet at the middle is:
The second body travels \(s/2\) at constant speed \(2v\) in time \(t = s/(4v)\). For the first body: \(s/2 = vt + \frac{1}{2}at^2\). Substituting \(t\) gives \(s/2 = s/4 + a s^2 / (32v^2)\), which simplifies to \(a = 8v^2/s\).
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 \(t = \):
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 \(t = 3\text{ s}\).
A particle is moving in vertical plane (x-y plane) such that its trajectory is given by the equation, \(y = x – \frac{x^2}{80}\), where \(x\) & \(y\) are in metre. For this particle match column-I with column-II and tick the correct option.
**Column I**:
A. Horizontal range (in m)
B. Angle of projection with horizontal (degree)
C. Maximum height gained by particle (m)
D. Speed of projection (\(\text{m s}^{-1}\))
**Column II**:
(P) 45
(Q) 80
(R) 20
(S) \(20\sqrt{2}\)
Comparing the trajectory to \(y = xtan\theta - \frac{gx^2}{2u^2cos^2\theta}\), we find \(tan\theta = 1⇒ \theta = 45^\circ\). Range at \(y = 0\) is \(x = 80\text{ m}\). Maximum height is \(H = \frac{Rtan\theta}{4} = 20\text{ m}\), and the initial speed \(u = 20\sqrt{2}\text{ m/s}\). Hence, A-Q, B-P, C-R, D-S.