JEE Physics Kinematics MCQs Set 09

Answer: (a) \( \sqrt{\pi^2+4}\ \text{m} \)

Explanation:
1. In half a revolution the centre moves forward \( \pi r=\pi \) m, and the marked point reaches the top, a height \( 2r=2 \) m above the ground.
2. Displacement \( =\sqrt{\pi^2+2^2}=\sqrt{\pi^2+4} \) m.

Teacher's Note:
1. The point moves along a cycloid, but displacement depends only on the start and end positions.
2. Use the horizontal and vertical offsets.

Answer: (a) 8 m

Explanation:
1. \( s=\int_0^2 4t\,dt=2t^2\big|_0^2 \).
2. \( s=8 \) m.

Teacher's Note:
1. A linearly increasing velocity means uniform acceleration p, so s = half p t squared.

Answer: (a) \( \dfrac{v_1-v_2}{v_2-v_3}=\dfrac{t_1+t_2}{t_2+t_3} \)

Explanation:
1. For uniform acceleration, the average velocity over an interval equals the instantaneous velocity at its midpoint.
2. The midpoints are separated by \( \tfrac{t_1+t_2}{2} \) and \( \tfrac{t_2+t_3}{2} \).
3. So \( v_1-v_2=-a\tfrac{t_1+t_2}{2} \), \( v_2-v_3=-a\tfrac{t_2+t_3}{2} \), and the ratio follows.

Teacher's Note:
1. Average velocity in uniform acceleration is the velocity at the middle instant of the interval.
2. The gaps between midpoints give the ratio.

Answer: (b) 3.06 m/s\(^2\) and 11.4 s

Explanation:
1. \( u=126\times\tfrac5{18}=35 \) m/s.
2. \( a=\dfrac{u^2}{2s}=\dfrac{1225}{400}=3.06 \) m/s\(^2\).
3. \( t=\dfrac{2s}{u}=\dfrac{400}{35}=11.4 \) s.

Teacher's Note:
1. Convert km/h to m/s first.
2. Then use v squared equals u squared minus 2as and the average-velocity formula for time.

Answer: (c) \( \dfrac2n-\dfrac1{n^2} \)

Explanation:
1. Distance in the nth second \( =\tfrac12a(2n-1) \).
2. Distance in n seconds \( =\tfrac12an^2 \).
3. Ratio \( =\dfrac{2n-1}{n^2}=\dfrac2n-\dfrac1{n^2} \).

Teacher's Note:
1. The nth-second distance is the difference of the distances in n and n minus 1 seconds.

Answer: (a) \( \dfrac{2(bp-aq)}{pq(p+q)} \)

Explanation:
1. Average velocities: \( a/p \) at the middle of the first interval and \( b/q \) at the middle of the second.
2. The time between these instants is \( \tfrac p2+\tfrac q2 \).
3. \( \text{accel}=\dfrac{b/q-a/p}{(p+q)/2}=\dfrac{2(bp-aq)}{pq(p+q)} \).

Teacher's Note:
1. Use the fact that average velocity over an interval equals the velocity at its midpoint under uniform acceleration.

Answer: (b) 4 m

Explanation:
1. The speed is \( |t-2| \), which changes form at t = 2 s.
2. \( s=\int_0^2(2-t)\,dt+\int_2^4(t-2)\,dt=2+2 \).
3. \( s=4 \) m.

Teacher's Note:
1. The distance is the area under the speed curve.
2. Split the integral at the point where t minus 2 changes sign.

Answer: (a) \( \sqrt{a^2+g^2}\ \text{and}\ g \)

Explanation:
1. With respect to the earth the ball falls freely, so its acceleration is g downward.
2. With respect to the bus: \( \vec a_{ball}-\vec a_{bus}=\vec g-\vec a \), whose magnitude is \( \sqrt{g^2+a^2} \).

Teacher's Note:
1. Once the ball leaves the bus only gravity acts on it.
2. The bus acceleration enters only in the relative acceleration.

Answer: (c) 35.35 km/h

Explanation:
1. For uniform acceleration the speed at the midpoint of the path satisfies \( v^2=\dfrac{u^2+w^2}{2} \).
2. \( v=\sqrt{\dfrac{900+1600}{2}}=\sqrt{1250}=35.35 \) km/h.

Teacher's Note:
1. The midpoint-of-distance speed is the root mean square of the end speeds, not their arithmetic mean.

Answer: (a) 22 m/s\(^2\)

Explanation:
1. \( x=t^4-t^2 \). Setting \( x=2 \): \( t^4-t^2-2=0 \), so \( t^2=2 \).
2. \( a=dv/dt=12t^2-2=24-2=22 \) m/s\(^2\).

Teacher's Note:
1. Integrate the velocity to find the position, solve for the time at x = 2, then differentiate for the acceleration.

Answer: (a) \( \dfrac{v_0}{\sqrt{2v_0^2kt+1}} \)

Explanation:
1. \( \int_{v_0}^{v}v^{-3}dv=-k\int_0^t dt \), so \( -\dfrac1{2v^2}+\dfrac1{2v_0^2}=-kt \).
2. \( \dfrac1{v^2}=\dfrac1{v_0^2}+2kt \).
3. \( v=\dfrac{v_0}{\sqrt{1+2kv_0^2t}} \).

Teacher's Note:
1. Separate the variables and integrate between the limits.
2. Invert the result for v.

Answer: (b) 18.75 m

Explanation:
1. Stopping distance \( =ut_0+\dfrac{u^2}{2a} \). So \( 10=10t_0+\dfrac{50}{a} \) and \( 30=20t_0+\dfrac{200}{a} \).
2. Subtracting twice the first from the second: \( 10=\dfrac{100}{a} \), so \( a=10 \) m/s\(^2\) and \( t_0=0.5 \) s.
3. For 15 m/s: \( 15\times0.5+\dfrac{225}{20}=7.5+11.25=18.75 \) m.

Teacher's Note:
1. Stopping distance has two parts: distance during reaction time and braking distance.
2. Two data sets give two unknowns.

Answer: (c) \( x\gt 0,\ v\lt 0,\ a\gt 0 \)

Explanation:
1. The lift is above the ground, so \( x\gt 0 \); it is moving down, so \( v\lt 0 \).
2. It is about to stop at the 4th floor, so its downward speed is decreasing: the acceleration is directed upwards, \( a\gt 0 \).

Teacher's Note:
1. A slowing downward motion has acceleration opposite to the velocity, which is upward here.

Answer: (a) \( a_2=2a_1 \)

Explanation:
1. Case 1: \( u^2=2a_1(4)+2a_2(1) \). Case 2: \( u^2=2a_2(2)+2a_1(2) \).
2. Equating: \( 8a_1+2a_2=4a_2+4a_1 \), so \( 4a_1=2a_2 \).
3. \( a_2=2a_1 \).

Teacher's Note:
1. Both bullets lose all their kinetic energy, so the two equations for u squared can be equated.

Answer: (a) \( -\dfrac{20}{3}\ \text{m/s}^2 \)

Explanation:
1. \( v=20-\tfrac23 s \), so at s = 15 m, \( v=10 \) m/s and \( dv/ds=-\tfrac23 \).
2. \( a=v\dfrac{dv}{ds}=10\times\left(-\dfrac23\right)=-\dfrac{20}{3} \) m/s\(^2\).

Teacher's Note:
1. When velocity is given as a function of displacement, use a = v dv/ds.

Answer: (a) 105 m/s

Explanation:
1. Bullet speed relative to ground \( =150+30\times\tfrac5{18}=158.3 \) m/s.
2. Car speed \( =192\times\tfrac5{18}=53.3 \) m/s.
3. Relative speed \( =158.3-53.3=105 \) m/s.

Teacher's Note:
1. The damaging speed is the relative speed of the bullet with respect to the car.

Answer: (a) 3/5

Explanation:
1. Let the maximum speed be v. The first and last stages have average speed v/2.
2. Times: \( \tfrac{2s}{v},\ \tfrac{2s}{v},\ \tfrac{6s}{v} \), total \( \tfrac{10s}{v} \).
3. Average velocity \( =\dfrac{6s}{10s/v}=0.6v \). Ratio \( =3/5 \).

Teacher's Note:
1. Find each stage's time using its average speed, then divide total distance by total time.

Answer: (a) \( \dfrac{c-2b+a}{n^2} \)

Explanation:
1. For uniform acceleration, the second difference of positions at equal time steps is \( an^2 \).
2. \( (c-b)-(b-a)=a_{cc}n^2 \).
3. \( a_{cc}=\dfrac{c-2b+a}{n^2} \).

Teacher's Note:
1. Equal time gaps give a constant second difference of position, equal to acceleration times the square of the gap.

Answer: (c) 56.6 s

Explanation:
1. Accelerate to speed v then brake: \( \dfrac{v^2}{2(1)}+\dfrac{v^2}{2(3)}=1200 \), so \( \tfrac23v^2=1200 \) and \( v=42.4 \) m/s.
2. Time \( =\dfrac v1+\dfrac v3=42.4+14.1 \).
3. \( t\approx56.6 \) s.

Teacher's Note:
1. The quickest trip uses full acceleration then full braking with no cruising.
2. Find the peak speed from the total distance.

Answer: (a) \( v_0t+\dfrac16bt^3 \)

Explanation:
1. \( v=v_0+\int_0^t bt\,dt=v_0+\tfrac12bt^2 \).
2. \( x=\int_0^t v\,dt=v_0t+\tfrac16bt^3 \).

Teacher's Note:
1. Integrate the acceleration once for velocity, with the initial velocity as the constant, and once more for displacement.