A body starts from rest, what is the ratio of the distance travelled by the body during the \(4^{\text{th}}\) and \(3^{\text{rd}}\) second?
(1993)
1. 7/5
2. 5/7
3. 7/3
4. 3/7
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Concept: Distance covered in the \(n^{text{th}}\) second for uniformly accelerated motion.
Formula: \(S_n = u + \frac{a}{2}(2n - 1)\). Since it starts from rest, \(u=0\).
Solution: \(S_4 = \frac{a}{2}(2(4) - 1) = \frac{7a}{2}\), \(S_3 = \frac{a}{2}(2(3) - 1) = \frac{5a}{2}\). Ratio \(S_4:S_3 = 7a/2 : 5a/2 = 7:5\).
If a car at rest accelerates uniformly to a speed of \(144 \text{ km/h}\) in \(20 \text{ sec}\), it covers a distance of:
(1997)
1. \(1440 \text{ cm}\)
2. \(2980 \text{ cm}\)
3. \(20 \text{ m}\)
4. \(400 \text{ m}\)
View Answer
Given (u=0), \(v = 144 \text{ km/h} = 144 \times \frac{5}{18} = 40 \text{ m/s}\), \(t = 20 \text{ s}\). Using \(S = \frac{u+v}{2}t), we get \(S = \frac{0+40}{2} \times 20 = 20 \times 20 = 400 \text{ m}\).
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. \(v_0 t + \frac{bt^3}{3}\)
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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 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)
1. 160 Km
2. 180 Km
3. 100 Km
4. 120 Km
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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)
1. \(\frac{(\alpha^2 - \beta^2)t}{\alpha\beta}\)
2. \(\frac{(\alpha^2 + \beta^2)t}{\alpha\beta}\)
3. \(\frac{(\alpha + \beta)t}{\alpha\beta}\)
4. \(\frac{\alpha\beta t}{\alpha + \beta}\)
View Answer
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)
1. \(-2n\beta^2 x^{-4n-1}\)
2. \(-2n\beta^2 x^{-2n+1}\)
3. \(-2n\beta^2 e^{-4n+1}\)
4. \(-2n\beta^2 x^{-2n-1}\)
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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)
1. \(24 m s^{-2}\)
2. (Zero)
3. \(6 m s^{-2}\)
4. \(12 m s^{-2}\)
View Answer
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 \(-a = 12 \text{ m/s}^2\).