JEE Physics Motion In A Straight Line MCQs Set 08

Answer: (d) \( \frac{1}{x^3} \)

Explanation:
1. Differentiate \( x^2 = t^2 + 1 \): \( 2x\,v = 2t \), so \( v = t/x \)
2. Differentiate again: \( a = \frac{1}{x} - \frac{t}{x^2}v = \frac{1}{x} - \frac{t^2}{x^3} \)
3. Combine: \( a = \frac{x^2 - t^2}{x^3} = \frac{1}{x^3} \) since \( x^2 - t^2 = 1 \). Answer: (d)

Teacher's Note:
1. Eliminate the unknown function by differentiating the given relation twice.
2. Use the relation itself to simplify at the end.

Answer: (b) Motion towards detector with constant velocity

Explanation:
1. Between t = 6 s and t = 8 s the graph is a straight line
2. Its position falls from about 1.2 m to 0 m, so the object comes closer to the detector
3. A straight line on a position-time graph means constant slope, that is constant velocity. Answer: (b)

Teacher's Note:
1. Slope of a position-time graph is velocity; a straight segment means constant velocity.
2. Decreasing position relative to the detector means motion towards it.

Answer: (b) \( -2 \)

Explanation:
1. \( a = \frac{dv}{dt} = \frac{d}{dt}(10 - 2t) = -2 \). Answer: (b)

Teacher's Note:
1. Acceleration is the time derivative of velocity.
2. The negative sign shows that the velocity decreases with time.

Answer: (c) 3 m/s²

Explanation:
1. Let \( a = a_0 + kt \). Then \( v = a_0 t + \frac{k t^2}{2} \) since \( v = 0 \) at \( t = 0 \)
2. At \( t = 2 \): \( 2a_0 + 2k = 6 \), so \( a_0 + k = 3 \)
3. At \( t = 4 \): \( 4a_0 + 8k = 20 \), so \( a_0 + 2k = 5 \)
4. Hence \( k = 2 \) and \( a_0 = 1 \)
5. At \( t = 1 \): \( a = 1 + 2(1) = 3 \) m/s². Answer: (c)

Teacher's Note:
1. Linear variation means a = a0 + kt, so v is quadratic in t.
2. Two velocity readings give two equations for the two constants.

Answer: (a) a straight line of positive slope cutting the \( t \)-axis at \( t = 1 \) and the \( y_2 \)-axis at \( y_2 = -2 \)

Explanation:
1. \( \frac{dy}{dt} = t^2 - 2t - 3 \)
2. \( y_2 = \frac{d^2y}{dt^2} = 2t - 2 \)
3. This is a straight line of slope +2 with \( y_2 = -2 \) at \( t = 0 \) and \( y_2 = 0 \) at \( t = 1 \). Answer: (a)

Teacher's Note:
1. Differentiate twice and read the intercepts from the linear result.
2. Slope sign and both intercepts must match for a line to be the right graph.

Answer: (a) \( a \propto v^3 \)

Explanation:
1. \( \frac{dt}{dx} = 2px + q = \frac1v \), so \( v = \frac{1}{2px+q} \)
2. \( a = v\frac{dv}{dx} = v \cdot \frac{-2p}{(2px+q)^2} = -2p\,v^3 \)
3. So \(
4. a
5. \propto v^3 \). Answer: (a)

Teacher's Note:
1. Write a as v dv/dx so that x is eliminated.
2. The constant factor -2p does not change the proportionality.

Answer: (b) \( -k e^{-bt} \)

Explanation:
1. \( v = \frac{dx}{dt} = \frac{k}{b^2}\cdot b e^{-bt} = \frac{k}{b}e^{-bt} \)
2. \( a = \frac{dv}{dt} = \frac{k}{b}(-b)e^{-bt} = -k e^{-bt} \). Answer: (b)

Teacher's Note:
1. Differentiate twice, and remember each differentiation of e^(-bt) brings a factor of -b.
2. The negative acceleration shows the velocity decays.

Answer: (a) 6 s

Explanation:
1. \( v = \frac{dx}{dt} = \frac1x \), so \( x\,dx = dt \)
2. Integrate: \( t = \int_2^4 x\,dx = \frac{4^2 - 2^2}{2} = 6 \) s. Answer: (a)

Teacher's Note:
1. Separate the variables before integrating.
2. The velocity stays finite and positive on the way, so x = 4 m is reached.