JEE Physics Motion In A Straight Line MCQs Set 17

Answer: (b) With respect to the frame of reference attached to the person sitting in the car, the car is at rest

Explanation:
1. Rest or motion is relative to the chosen frame of reference.
2. A moving car changes its position with respect to the ground and with respect to a person standing outside, so statements 1 and 3 are false.
3. A person sitting in the car does not change position relative to the car, so the car is at rest in that frame.

Teacher's Note:
1. Always ask 'with respect to what?' before calling a body moving or at rest.
2. An observer moving with the body sees it at rest.

Answer: (a) Graph (a)

Explanation:
1. An object at rest does not change its position, so x stays constant as t increases.
2. The slope of an x-t graph is the velocity, and a horizontal line has zero slope.
3. Graph (a) is the horizontal line.

Teacher's Note:
1. Slope of x-t graph = velocity.
2. A straight sloping line means uniform velocity; a horizontal line means rest.

Answer: (a) motion along the Z-axis

Explanation:
1. Only the z-coordinate changes (3 to 4); x and y stay the same.
2. So the displacement is along the Z-axis.

Teacher's Note:
1. Compare coordinates one by one to find the direction of displacement.
2. A change in position means the object is not at rest.

Answer: (c) Graph (c)

Explanation:
1. Constant velocity means the position changes by equal amounts in equal intervals of time.
2. Hence x is a linear function of t, a sloping straight line through the origin, as in graph (c).
3. A horizontal line is a body at rest, a curved line shows changing velocity.

Teacher's Note:
1. Uniform motion gives a straight x-t line whose slope is the velocity.
2. Curves in an x-t graph imply acceleration.

Answer: (a) Graph (a)

Explanation:
1. 0 to 10 s: speed increases, so the x-t graph curves upwards (slope increasing).
2. 10 s to 16 s: uniform speed gives a straight line segment.
3. 16 s to 20 s: speed decreases to zero, so the curve flattens and its slope becomes zero at t = 20 s, x = 296 m.
4. Only graph (a) has all three features; graph (b) is a single straight line and graph (c) keeps getting steeper.

Teacher's Note:
1. Read the slope of the x-t graph piece by piece: increasing, constant, decreasing to zero.
2. The car ends at rest, so the graph must end with a horizontal tangent.

Answer: (c) 8h/9 metres from the ground

Explanation:
1. \( h=\tfrac12gT^2 \).
2. In time T/3 the ball drops \( \tfrac12g\frac{T^2}{9}=\frac{h}{9} \).
3. Its height above the ground is \( h-\frac{h}{9}=\frac{8h}{9} \).

Teacher's Note:
1. The distance dropped is proportional to the square of the time, so one third of the time gives one ninth of the drop.
2. Remember to subtract from h, since the question asks for the height above the ground.

Answer: (a) 2.5 h

Explanation:
1. Take A as origin. Position of B relative to A: \( \vec r=(0,-100)+(20,20)t \) (relative velocity = 20 N + 20 E).
2. Distance squared: \( d^2=(20t)^2+(20t-100)^2 \).
3. Set derivative to zero: \( 2(20)(20t)+2(20)(20t-100)=0\Rightarrow 800t=2000 \), so t = 2.5 h.

Teacher's Note:
1. Use relative velocity to turn a two-body problem into a one-body problem.
2. The distance is shortest when relative position is perpendicular to relative velocity.

Answer: (b) 10 m/s

Explanation:
1. Average velocity \( =\frac{\Delta x}{\Delta t} \).
2. \( =\frac{30-10}{7-5}=10\ \text{m/s} \).

Teacher's Note:
1. Average velocity is displacement divided by time, which is the slope of the chord joining the two points on the x-t graph.
2. The shape of the curve in between does not matter.

Answer: (a) 1 : √3

Explanation:
1. Velocity equals the slope of the x-t graph: \( v=\tan\theta \).
2. B makes 30° + 15° = 45° with the time axis.
3. \( v_A:v_B=\tan30^\circ:\tan45^\circ=\frac{1}{\sqrt3}:1=1:\sqrt3 \).

Teacher's Note:
1. Slope = tan of the angle with the time axis, not the angle itself.
2. Add the 15° to the angle of A to get the angle of B.

Answer: (a) Graph (a)

Explanation:
1. The velocity starts at a positive value and decreases at a constant rate g, so the graph is a straight line with negative slope.
2. It becomes zero at the highest point and then negative while the particle comes down; the line crosses the time axis.
3. Only graph (a) is a straight descending line crossing the axis.

Teacher's Note:
1. Constant acceleration gives a straight line in the v-t graph.
2. The slope of the line is -g when upward is positive.

Answer: (b) Graph (b)

Explanation:
1. Positive velocity means x increases with time, i.e. the x-t graph has a positive slope.
2. Graph (a) has zero slope (rest) and graphs (c) and (d) have negative or zero slope.
3. Graph (b) is the only one with positive slope.

Teacher's Note:
1. Velocity is the slope of the x-t graph.
2. Check the sign of the slope in each portion of the graph.

Answer: (b) \( \sqrt3\ \text{m/s} \)

Explanation:
1. The graph is a straight line, so the velocity is constant and equals the slope.
2. \( v=\tan60^\circ=\sqrt3\ \text{m/s} \).

Teacher's Note:
1. Slope of x-t graph = tan of the angle with the time axis.
2. For a straight line the average and instantaneous velocities are equal.

Answer: (a) zero

Explanation:
1. At t = t₁ the graph is at its maximum, so the tangent is horizontal.
2. Slope zero means instantaneous velocity zero.

Teacher's Note:
1. Instantaneous velocity is the slope of the tangent to the x-t curve.
2. At a maximum or minimum of x, the velocity is zero.

Answer: (b) \( -2n\beta^2x^{-4n-1} \)

Explanation:
1. \( a=\frac{dv}{dt}=v\frac{dv}{dx} \).
2. \( \frac{dv}{dx}=-2n\beta x^{-2n-1} \).
3. \( a=\beta x^{-2n}\cdot(-2n\beta x^{-2n-1})=-2n\beta^2x^{-4n-1} \).

Teacher's Note:
1. When v is given as a function of x, use a = v dv/dx.
2. Powers of x add when multiplied: -2n + (-2n - 1) = -4n - 1.

Answer: (a) Must be zero

Explanation:
1. Distance is the total length of the path and is never less than the magnitude of displacement.
2. If the distance is zero the particle has not moved at all, so the displacement is zero.

Teacher's Note:
1. Magnitude of displacement is always less than or equal to the distance.
2. The converse is not true: zero displacement does not imply zero distance.

Answer: (d) Zero

Explanation:
1. Average velocity = total displacement / total time.
2. For the round trip the car returns to its starting point, so the displacement is zero.
3. Hence the average velocity is zero (the average speed would be 34.3 km/h).

Teacher's Note:
1. Do not confuse average velocity with average speed.
2. Zero displacement over any time interval gives zero average velocity.

Answer: (c) \( -7.5\ \text{m s}^{-2} \)

Explanation:
1. \( a=\frac{v-u}{t}=\frac{15-30}{2}=-7.5\ \text{m s}^{-2} \).

Teacher's Note:
1. A decrease of velocity gives a negative acceleration.
2. Keep the sign: this is a retardation of 7.5 m/s².

Answer: (c) 1 : 3 : 5

Explanation:
1. From rest, distances in successive equal intervals are in the ratio 1 : 3 : 5.
2. Check with g = 10: 20 m, 60 m, 100 m (total 180 m, less than 200 m), ratio 1 : 3 : 5.

Teacher's Note:
1. Galileo's law of odd numbers: 1 : 3 : 5 : 7 ... for equal successive intervals from rest.
2. The body must still be in the air after 6 s: 180 m < 200 m.

Answer: (a) \( \sqrt{2ax} \)

Explanation:
1. \( v^2=u^2+2ax \) with u = 0.
2. \( v=\sqrt{2ax} \).

Teacher's Note:
1. This is the time-independent equation of motion.
2. Check units: (m/s²·m)^(1/2) = m/s.

Answer: (a) \( (3x^2-4x)(6x-4) \)

Explanation:
1. \( a=v\frac{dv}{dx} \).
2. \( \frac{dv}{dx}=6x-4 \).
3. \( a=(3x^2-4x)(6x-4) \).

Teacher's Note:
1. When v depends on x, use a = v dv/dx (chain rule).
2. Do not differentiate with respect to t directly.

Answer: (b) Graph (b)

Explanation:
1. While both are in the air, \( y_2-y_1=(40-10)t=30t \), a straight line (same acceleration, so it cancels).
2. The first stone hits the ground when \( 10t-5t^2=-240 \), i.e. t = 8 s. At that time \( y_2-y_1=240\ \text{m} \).
3. After 8 s the first stone stays on the ground (y₁ = −240) and the second continues: \( y_2-y_1=40t-5t^2+240 \), a downward-opening parabola that reaches zero at t = 12 s when stone 2 lands.
4. Straight line up to 8 s followed by a curve that bends downwards (concave) to 12 s: graph (b).

Teacher's Note:
1. Relative motion of two bodies with the same acceleration is uniform motion until one of them stops.
2. After the first stone lands, the relative motion is just that of the second stone, so the curve is a parabola.

Answer: (d) \( \frac{5v_1v_2}{3v_1+2v_2} \)

Explanation:
1. Total distance d. Time \( =\frac{2d/5}{v_1}+\frac{3d/5}{v_2}=\frac{d(2v_2+3v_1)}{5v_1v_2} \).
2. \( v_{av}=\frac{d}{t}=\frac{5v_1v_2}{3v_1+2v_2} \).

Teacher's Note:
1. Average speed = total distance / total time, never the mean of the speeds unless times are equal.
2. Find the time for each part and add them.

Answer: (c) Graph (c)

Explanation:
1. From 0 to 3 s the velocity stays constant at 30 m/s: a horizontal line.
2. From 3 s to 5 s it rises uniformly to 40 m/s: a sloping straight line.
3. Graph (a) shows the velocity falling first, and graph (b) shows the velocity rising from the start; only graph (c) matches.

Teacher's Note:
1. Uniform velocity is a horizontal v-t line and uniform acceleration is a sloping straight line.
2. Check the time at which the slope changes: here it is 3 s.

Answer: (d) All of these

Explanation:
1. With upward positive, a = −9.8 m/s² and u = 0.
2. v = at = −9.8t; y = ½at² = −4.9t²; v² = 2ay = −19.6y.
3. All three equations are correct.

Teacher's Note:
1. Use the three equations of motion with a = −g when upward is positive.
2. The negative signs show that the object moves downwards.

Answer: (c) A and B reach home at the same time

Explanation:
1. Q is farther from O than P, so B lives farther from the school (statement 1 is false).
2. B's line starts on the time axis later than A's, so B starts later (statement 2 is false).
3. Both graphs end on the same vertical dashed line, so they reach home at the same time (statement 3 is true).
4. The lines intersect only once, so no 'twice' (statement 4 is false).

Teacher's Note:
1. Read off starting times from the points where the graphs leave the time axis.
2. Points where graphs cross show equal positions at the same time.

Answer: (b) \( d_s=\frac{v_0^2}{2a} \)

Explanation:
1. \( 0=v_0^2-2ad_s \).
2. \( d_s=\frac{v_0^2}{2a} \).

Teacher's Note:
1. Final velocity is zero at the stopping point.
2. Stopping distance grows as the square of the speed.

Answer: (a) Graph (a)

Explanation:
1. Speed is the slope of the x-t line. A must have the larger slope.
2. In graph (b) the line of B is steeper (in magnitude) than that of A, and in graph (c) the line labelled B is steeper than A, so both give vB > vA.
3. In graph (a) A (starting at 10 m and meeting B at 80 m at t = 3 s) is steeper than B (starting at 40 m): slope of A = 70/3, slope of B = 40/3, so vA > vB. Graph (a) is correct.

Teacher's Note:
1. Compare the slopes of the two lines, not their starting positions.
2. The meeting point (intersection) shows equal positions, not equal speeds.

Answer: (c) 95 m/s

Explanation:
1. First 4 s: \( 100=4u+8a \). First 6 s: \( 220=6u+18a \).
2. From the first, u = 25 − 2a. Substitute: 220 = 150 − 12a + 18a, so \( a=\frac{35}{3} \) m/s² and \( u=\frac{5}{3} \) m/s.
3. \( v=u+at=\frac53+\frac{35}{3}\times8=95\ \text{m/s} \).

Teacher's Note:
1. Two equations in u and a come from two different time intervals measured from the start.
2. Velocity at the end of the 8th second means t = 8 s.

Answer: (c) 160 km

Explanation:
1. Uniform acceleration: average velocity \( =\frac{20+60}{2}=40\ \text{km/h} \).
2. Distance = 40 × 4 = 160 km.

Teacher's Note:
1. With uniform acceleration the average velocity is the mean of the end values.
2. Units can be used as given (km and h) when the answer is in km.

Answer: (c) 12 seconds

Explanation:
1. \( v=\frac{dy}{dt}=-\frac43t+16 \).
2. At rest v = 0: \( t=\frac{16\times3}{4}=12\ \text{s} \).

Teacher's Note:
1. The body comes to rest when its velocity is zero.
2. Differentiate the displacement once to get the velocity.