The ratio of the distance traveled by a freely falling body in the \(1^{text{st}}\,\text{ }2^{text{nd}}\,\text{ }3^{text{rd}}\) and \(4^{text{th}}\) second:
(2022)
Concept: Galileo's law of odd numbers for free fall.
Formula: Distance in \(n^{text{th}}\) second is \(S_n = \frac{g}{2}(2n - 1)\).
Solution: \(S_1:S_2:S_3:S_4 = (2(1)-1):(2(2)-1):(2(3)-1):(2(4)-1) = 1:3:5:7\).
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)
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)
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)
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]
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 particle starts from rest with constant acceleration. The ratio of average velocity to the time average velocity is:
(1999)
For constant acceleration starting from rest, the displacement \(S = \frac{1}{2}at^2\). The average velocity over time (t) is \(v_{avg} = \frac{S}{t} = \frac{1}{2}at\). The final velocity at time (t) is (v = at). The question asks for the ratio of average velocity to final velocity (assuming "time average velocity" refers to final velocity). Ratio is \(\frac{(1/2)at}{at} = \frac{1}{2}\).
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)
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 car accelerates from rest at a constant rate \(\alpha\) for some time after which it decelerates at a constant rate \(\beta\) and comes to rest. If total time elapsed is t, then maximum velocity acquired by car will be:
(1994)
Let (v_{max}) be the maximum velocity. Time to accelerate: \(t_1 = \frac{v_{max}}{\alpha}). Time to decelerate: (t_2 = \frac{v_{max}}{\beta}). Total time (t = t_1 + t_2 = \frac{v_{max}}{alpha} + \frac{v_{max}}{beta} = v_{max}\left(\frac{1}{alpha} + \frac{1}{\beta}\right) = v_{max}\left(\frac{\beta + \alpha}{\alpha\beta}\right)). Solving for \(v_{max}): (v_{max} = \frac{\alpha\beta t}{\alpha + \beta}).
A particle of unit mass undergoes one dimensional motion such that its velocity varies according to \(v(x) = \beta x^{-2n}\) where \(\beta\) and (n) are constants and (x) is the position of the particle. The acceleration of the particle as a function of (x), is given by:
(2015)
Given \(v = \beta x^{-2n}\). Acceleration \(a = v \frac{dv}{dx}\). First find \(\frac{dv}{dx} = \beta (-2n)x^{-2n-1}\). Then \(a = (\beta x^{-2n})(-2n\beta x^{-2n-1}\) = \(-2n\beta^2 x^{-4n-1}\).
The motion of a particle along a straight line is described by equation: \\(x = 8 + 12t – t^3\) where (x) is in metre and (t) in second. The retardation of the particle when its velocity becomes zero, is:
(2012 Pre)
Given \(x = 8 + 12t - t^3\). Velocity \(v = \frac{dx}{dt} = 12 - 3t^2\). Acceleration \(a = \frac{dv}{dt} = -6t\). When (v=0), \(12 - 3t^2 = 0 \Rightarrow t^2 = 4 \Rightarrow t = 2 \text{ s}\). At \(t=2 \text{ s}\), \(a = -6(2) = -12 \text{ m/s}^2\). Retardation is \(-a = 12 \text{ m/s}^2\).