JEE Physics Motion In A Straight Line MCQs Set 05

Answer: (d) \( \frac v{\sqrt2} \)

Explanation:
1. Displacement \( =\sqrt{(vt)^2+(vt)^2}=\sqrt2vt \).
2. Average velocity \( =\frac{\sqrt2vt}{2t}=\frac v{\sqrt2} \).

Teacher's Note:
1. Average velocity is total displacement over total time.

Answer: (d) 16 m

Explanation:
1. \( v=12-3t^2=0\Rightarrow t=2 \) s.
2. \( x(0)=40 \) m and \( x(2)=40+24-8=56 \) m, so the distance is \( 56-40=16 \) m.

Teacher's Note:
1. Find when the velocity is zero, then the position change.

Answer: (b) \( \frac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1} \)

Explanation:
1. Time \( =\frac d{3v_1}+\frac d{3v_2}+\frac d{3v_3} \).
2. Average velocity \( =\frac d{\text{time}}=\frac3{\frac1{v_1}+\frac1{v_2}+\frac1{v_3}}=\frac{3v_1v_2v_3}{v_1v_2+v_2v_3+v_3v_1} \).

Teacher's Note:
1. This is the harmonic mean of the three velocities.

Answer: (c) equal to or less than one

Explanation:
1. The magnitude of displacement never exceeds the path length.
2. So the ratio is at most 1, with equality for motion along a straight line without turning back.

Teacher's Note:
1. The shortest path between two points is a straight line.

Answer: (c) \( 11:7\sqrt2 \)

Explanation:
1. Distance \( =\frac{\pi r}2=\frac{11r}7 \).
2. Displacement \( =\) chord \( =r\sqrt2 \).
3. Ratio \( =\frac{11}{7\sqrt2} \), i.e. \( 11:7\sqrt2 \).

Teacher's Note:
1. Distance is the arc length and displacement is the chord.

Answer: (d) directly proportional to the square root of the displacement

Explanation:
1. \( s=\frac12at^2\Rightarrow t=\sqrt{\frac{2s}a} \).
2. So t is proportional to the square root of s.

Teacher's Note:
1. Use s = (1/2) a t squared.

Answer: (c) the acceleration of the particle is 2a

Explanation:
1. \( v=u+2a(t-2) \), so the initial velocity (at t = 0) is \( u-4a \), not u.
2. The acceleration is \( \frac{dv}{dt}=2a \).
3. At \( t=2 \), \( x=0 \), so the particle is at the origin.

Teacher's Note:
1. Differentiate twice and check each statement.

Answer: (b) the body is sure to slow down

Explanation:
1. An opposite force produces an acceleration opposite to the velocity, so the speed decreases.
2. The body may or may not stop and reverse, so (a) is not necessary; (c) is false.

Teacher's Note:
1. Opposite force means deceleration.

Answer: (b) \( a+2bt-3ct^2 \)

Explanation:
1. \( v=\frac{dx}{dt}=a+2bt-3ct^2 \).

Teacher's Note:
1. Differentiate x with respect to t.

Answer: (c) \( \frac{2v_1v_2}{v_1+v_2} \)

Explanation:
1. Time \( =\frac{d/2}{v_1}+\frac{d/2}{v_2} \).
2. Average speed \( =\frac{2v_1v_2}{v_1+v_2} \).

Teacher's Note:
1. Use total distance over total time.

Answer: (b) a parabola with its axis along the s axis (v squared proportional to s)

Explanation:
1. \( v^2=2as \), so \( v\propto\sqrt s \).
2. The graph is a parabola opening along the s axis, rising steeply at first and flattening.

Teacher's Note:
1. Use v squared = 2 a s.

Answer: (b) \( \frac{4u^2}g \)

Explanation:
1. The net displacement is h downwards: \( (3u)^2=u^2+2gh \).
2. \( 8u^2=2gh\Rightarrow h=\frac{4u^2}g \).

Teacher's Note:
1. Use v squared = u squared + 2 g h with the same g throughout.

Answer: (d) 180 m

Explanation:
1. The retardation is the same, so \( s\propto v^2 \).
2. \( s=20\times\left(\frac{30}{10}\right)^2=180 \) m.

Teacher's Note:
1. The distance is proportional to the square of the speed.

Answer: (a) \( \sqrt{\frac h{8g}} \)

Explanation:
1. The second stone's speed: \( u^2=2g(4h)=8gh \).
2. Relative motion: the gap h closes at speed u, so \( t=\frac hu=\frac h{\sqrt{8gh}}=\sqrt{\frac h{8g}} \).

Teacher's Note:
1. Use relative motion; gravity cancels.

Answer: (a) Zero

Explanation:
1. 140 s corresponds to \( 140/20=7 \) complete rounds.
2. After a whole number of rounds the athlete is back at the start, so the displacement is zero.

Teacher's Note:
1. Count the number of full rounds.

Answer: (a) 4 s

Explanation:
1. \( v=\frac{ds}{dt}=12t-3t^2=3t(4-t) \).
2. \( v=0 \) at \( t=0 \) and again at \( t=4 \) s.

Teacher's Note:
1. Set the velocity to zero and discard t = 0.

Answer: (c) non-uniform acceleration

Explanation:
1. \( a=\frac{dv}{dt}=0.2t \), which changes with time and is positive.
2. So the acceleration is non-uniform.

Teacher's Note:
1. Differentiate v with respect to t.

Answer: (c) \( \frac{3\pi}4 \)

Explanation:
1. Distance by P \( =\frac34\cdot2\pi r=\frac{3\pi r}2 \).
2. Distance by Q \( =r+r=2r \).
3. Ratio \( =\frac{3\pi r/2}{2r}=\frac{3\pi}4 \).

Teacher's Note:
1. Distance depends on the path, not just the end points.

Answer: (b) displacement

Explanation:
1. \( \int v\,dt \) is the displacement.
2. So the area under the v-t curve represents displacement.

Teacher's Note:
1. Area under v-t gives displacement.

Answer: (d) OAB and BC

Explanation:
1. The velocity is positive (upward motion) from O to B and negative (downward motion) from B to C.
2. So it goes up along OAB and comes down along BC.

Teacher's Note:
1. Check the sign of the velocity.

Answer: (b) 100 m

Explanation:
1. 72 km/h = 20 m/s.
2. \( s=\frac{0+20}2\times10=100 \) m.

Teacher's Note:
1. Use the average velocity.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. The straight line between two points is the shortest path, so the assertion is true.
2. The reason is also true, but it is a different fact and does not explain the assertion.

Teacher's Note:
1. Distance is the length of the path.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Average speed is path length over time and the magnitude of average velocity is displacement over time.
2. Since path length is at least the magnitude of displacement, the assertion follows from the reason.

Teacher's Note:
1. Divide both quantities by the same time.