JEE Physics Motion In A Straight Line MCQs Set 07

Answer: (c) \( t=\sqrt{t_1t_2} \)

Explanation:
1. With \( h \) the height: \( -h=ut_1-\frac12gt_1^2 \) and \( h=ut_2+\frac12gt_2^2 \).
2. Eliminating u gives \( h=\frac12gt_1t_2 \). Free fall: \( h=\frac12gt^2\), so \( t=\sqrt{t_1t_2} \).

Teacher's Note:
1. Eliminate the speed u between the two equations.

Answer: (b) exactly equal

Explanation:
1. In vacuum there is no air resistance, so both fall with acceleration g regardless of mass.
2. The times are exactly equal.

Teacher's Note:
1. Free fall is independent of mass.

Answer: (b) velocity zero, speed 10 m/s

Explanation:
1. Average speed \( =\frac{2\pi r}T=\frac{2\pi\times10}{6.28}=10 \) m/s.
2. Over a full lap the displacement is zero, so the average velocity is zero.

Teacher's Note:
1. Distance over time gives speed; displacement over time gives velocity.

Answer: (c) 7 m/s

Explanation:
1. Distance \( =15\times2+5\times8=30+40=70 \) m in 10 s.
2. Average speed \( =7 \) m/s.

Teacher's Note:
1. Average speed uses the distance, not the displacement.

Answer: (b) may or may not be zero

Explanation:
1. A particle can return to its starting point after moving along a path.
2. Then the displacement is zero but the distance is not.
3. If it does not move at all, both are zero, so the distance may or may not be zero.

Teacher's Note:
1. Displacement depends only on the end points.
2. Distance is the length of the path.

Answer: (c) 4 m/s

Explanation:
1. Let the maximum speed be v. Time to accelerate \( =\frac v2 \), time to stop \( =\frac v4 \).
2. \( \frac v2+\frac v4=3\Rightarrow v=4 \) m/s.

Teacher's Note:
1. Write the total time as the sum of the two stages.

Answer: (a) 8.0 m/s

Explanation:
1. For the second half, equal time intervals give an average speed of \( \frac{4.5+7.5}2=6 \) m/s.
2. Total distance d: time \( =\frac{d/2}{12}+\frac{d/2}{6}=\frac d{24}+\frac d{12}=\frac d8 \).
3. Average speed \( =\frac d{d/8}=8 \) m/s.

Teacher's Note:
1. Treat the second half as a single stretch with average speed 6 m/s.

Answer: (a) 9 m

Explanation:
1. \( s=\int_0^3kt\,dt=\frac{k}2(3)^2=\frac22\cdot9=9 \) m.

Teacher's Note:
1. Integrate velocity over time.

Answer: (a) 15 m/s

Explanation:
1. \( v=\int_0^52(t-1)\,dt=\left[t^2-2t\right]_0^5=25-10=15 \) m/s.

Teacher's Note:
1. Integrate from 0 to 5 s.

Answer: (a) \( \frac75 \)

Explanation:
1. \( s_n\propto(2n-1) \): \( s_4:s_3=7:5 \).

Teacher's Note:
1. Use the nth second formula.

Answer: (c) \( \frac a{3b} \)

Explanation:
1. \( v=2at-3bt^2 \) and acceleration \( =2a-6bt \).
2. Setting it to zero: \( t=\frac{a}{3b} \).

Teacher's Note:
1. Differentiate twice and set the result to zero.

Answer: (d) 41 %

Explanation:
1. \( p=mv=m\sqrt{2gh}\propto\sqrt h \).
2. New height \( 2h \): \( p'=\sqrt2p\approx1.414p \), a 41 % increase.

Teacher's Note:
1. Momentum is proportional to the square root of the height.

Answer: (a) 16 m/s\( ^2 \)

Explanation:
1. \( s_8=\frac a2(2\times8-1)=\frac{15a}2=120\Rightarrow a=16 \) m/s\( ^2 \).

Teacher's Note:
1. Use the nth second formula.

Answer: (d) 1 m/s\( ^2 \)

Explanation:
1. 72 km/h = 20 m/s.
2. \( a=\frac{v^2}{2s}=\frac{400}{400}=1 \) m/s\( ^2 \).

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

Answer: (c) 31.25 m

Explanation:
1. First stone: \( h=5T^2 \). Second stone falls \( h-20 \) in time \( T-1 \): \( h-20=5(T-1)^2 \).
2. Subtracting: \( 20=10T-5\Rightarrow T=2.5 \) s.
3. \( h=5\times6.25=31.25 \) m.

Teacher's Note:
1. Write both fall equations and subtract.

Answer: (b) 12 m/s\( ^2 \)

Explanation:
1. \( v=12-3t^2=0\Rightarrow t=2 \) s.
2. \( a=-6t=-12 \) m/s\( ^2 \) at t = 2 s, a retardation of 12 m/s\( ^2 \).

Teacher's Note:
1. Find the time first.

Answer: (b) 75 m/s

Explanation:
1. First ball: \( \frac12(10)(18)^2=1620 \) m.
2. Second ball (12 s of motion): \( 12v+\frac12(10)(144)=12v+720 \).
3. \( 12v+720=1620\Rightarrow v=75 \) m/s.

Teacher's Note:
1. Equate the distances covered at t = 18 s.

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

Explanation:
1. Relative to each other the bodies approach at speed u, so they meet after \( \frac hu \): the assertion is true.
2. In the last second of ascent the body covers \( \frac12g(1)^2=4.9 \) m, so the reason is true, but it has nothing to do with the meeting time.

Teacher's Note:
1. Gravity cancels in the relative motion.

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

Explanation:
1. A body thrown up or down is in free fall (only gravity acts) although its initial velocity is not zero.
2. Free fall is defined by the acceleration g, which explains why the initial velocity can be anything.

Teacher's Note:
1. Free fall is defined by the acceleration, not the velocity.

Answer: (c) A is true, but R is false.

Explanation:
1. With constant velocity all three quantities have the same magnitude, so the assertion is true.
2. If the average velocity over an interval is zero, the particle returns to its starting point, so it must reverse and its velocity must be zero at some instant. The reason is false.

Teacher's Note:
1. A continuous velocity must pass through zero when it changes sign.

Answer: (d) A is false, but R is true.

Explanation:
1. Speed is the magnitude of velocity and is never negative, so the assertion is false.
2. Motion in the negative direction gives negative velocity, not negative speed, so the reason is false too.

Teacher's Note:
1. Speed is a magnitude.