JEE Physics Motion In A Straight Line MCQs Set 12

Answer: (c) \(a+\dfrac{b^2}{3c}\)

Explanation:
1. \(v=a+2bt-3ct^2\), \(a_{cc}=2b-6ct\)
2. Zero acceleration at \(t=\dfrac{b}{3c}\)
3. \(v=a+\dfrac{2b^2}{3c}-\dfrac{b^2}{3c}=a+\dfrac{b^2}{3c}\)

Teacher's Note:
1. Solve for the time of zero acceleration first.
2. Substitute that time into the velocity expression.

Answer: (b) \(2\hat i+4\hat j\)

Explanation:
1. \(y=\dfrac{t^4}{8}\)
2. \(v_x=\dfrac{dx}{dt}=t=2\), \(v_y=\dfrac{dy}{dt}=\dfrac{t^3}{2}=4\)
3. \(\vec v=2\hat i+4\hat j\) m/s

Teacher's Note:
1. Express both coordinates in terms of t and differentiate each separately.
2. Components of velocity along x and y are independent.

Answer: (b) 16 m

Explanation:
1. \(s=\int_0^4 2t\,dt=t^2\Big|_0^4\)
2. \(=16\) m

Teacher's Note:
1. Distance is the area under the velocity-time graph.
2. Here the graph is a straight line through the origin, a triangle of area \(\tfrac12\times4\times8=16\).

Answer: (d) \(\dfrac{1}{\sqrt2}\) m/s² towards north-west

Explanation:
1. \(\Delta\vec v=10\hat j-10\hat i\), magnitude \(10\sqrt2\) m/s, pointing north-west
2. Average acceleration \(=\dfrac{10\sqrt2}{20}=\dfrac1{\sqrt2}\) m/s² towards north-west

Teacher's Note:
1. Average acceleration is change in velocity vector divided by time.
2. The change vector points from the tip of the initial vector to the tip of the final vector.

Answer: (d) 14 m/s²

Explanation:
1. \(v(2)=18\) m/s and \(v(5)=60\) m/s
2. Average acceleration \(=\dfrac{60-18}{5-2}=14\) m/s²

Teacher's Note:
1. Average acceleration uses the change in velocity over the interval.
2. Instantaneous acceleration \(4t\) would give a different value.

Answer: (a) \(v\propto t\)

Explanation:
1. \(\dfrac{dx}{\sqrt x}=\alpha\,dt\), so \(2\sqrt x=\alpha t\) and \(x=\dfrac{\alpha^2t^2}{4}\)
2. \(v=\dfrac{dx}{dt}=\dfrac{\alpha^2t}{2}\), so \(v\propto t\)

Teacher's Note:
1. Separate the variables and integrate to get x(t).
2. The acceleration is constant, \(\alpha^2/2\).

Answer: (c) -8 m/s²

Explanation:
1. \(a=v\dfrac{dv}{dx}=\dfrac12\dfrac{d(v^2)}{dx}\)
2. \(v^2=180-16x\), so \(a=\dfrac12(-16)=-8\) m/s²

Teacher's Note:
1. When velocity is given in terms of position, use \(a=v\,dv/dx\).
2. The acceleration is constant here, as \(v^2\) is linear in x.

Answer: (a) 1 s

Explanation:
1. \(\vec v=(2t-4)\hat i+2t\hat j\) and \(\vec a=2\hat i+2\hat j\)
2. Perpendicular means \(\vec v\cdot\vec a=0\): \(2(2t-4)+2(2t)=0\)
3. \(8t=8\), so t = 1 s

Teacher's Note:
1. Two vectors are perpendicular when their dot product is zero.
2. Differentiate r once for velocity and twice for acceleration.

Answer: (d) \(\dfrac{7}{25}(3\hat i+4\hat j)\)

Explanation:
1. Tangential acceleration \(=\vec a\cdot\hat v\) along \(\hat v\)
2. \(\hat v=\dfrac{6\hat i+8\hat j}{10}\), so \(\vec a\cdot\hat v=\dfrac{6+8}{10}=1.4\)
3. Vector \(=1.4\times\dfrac{3\hat i+4\hat j}{5}=\dfrac{7}{25}(3\hat i+4\hat j)\)

Teacher's Note:
1. The component of acceleration along the velocity changes the speed.
2. The perpendicular component only turns the velocity.

Answer: (b) 400 m

Explanation:
1. 144 km/h = 40 m/s
2. \(s=\dfrac{(u+v)}{2}t=\dfrac{0+40}{2}\times20\)
3. \(=400\) m

Teacher's Note:
1. Convert km/h to m/s by multiplying by 5/18.
2. For uniform acceleration from rest, distance = average speed times time.

Answer: (b) 320 m

Explanation:
1. Same force means same retardation, so \(s\propto u^2\)
2. \(s=20\times\left(\dfrac{40}{10}\right)^2=320\) m

Teacher's Note:
1. Stopping distance is proportional to the square of the initial speed.
2. Doubling the speed quadruples the stopping distance.

Answer: (b) 312.5 m

Explanation:
1. Phase 1: \(s_1=\tfrac12\times5\times25=62.5\) m and \(v=25\) m/s
2. Phase 2: \(s_2=25\times10=250\) m
3. Total \(=312.5\) m

Teacher's Note:
1. Split the motion into the accelerated and uniform phases.
2. The speed at the end of phase 1 is the constant speed of phase 2.

Answer: (c) 6 m/s

Explanation:
1. \(a=\dfrac{20-10}{5}=2\) m/s²
2. Velocity 2 s earlier \(=10-2\times2=6\) m/s

Teacher's Note:
1. Constant acceleration lets you move backwards in time with the same formula.
2. Subtract a times the time interval.

Answer: (a) A body can have zero velocity and still be accelerated

Explanation:
1. At the highest point of a vertical throw the velocity is zero but the acceleration is g, so (a) is true
2. Constant velocity means constant magnitude, so speed cannot vary; (b) is false
3. Uniform circular motion has constant speed with varying velocity; (c) is false
4. A projectile has constant acceleration but changing direction of velocity; (d) is false

Teacher's Note:
1. Velocity is a vector, so direction changes count as changes in velocity.
2. Zero velocity at one instant does not mean zero acceleration.

Answer: (c) 5 s

Explanation:
1. Let the total time be T and take \(g=10\) m/s²
2. Distance in last second \(=\tfrac12g\left[T^2-(T-1)^2\right]=\tfrac12g(2T-1)\)
3. Distance in first 3 s \(=\tfrac12g(9)\)
4. Equating, \(2T-1=9\), so T = 5 s

Teacher's Note:
1. Write the distance in the last second as the difference of two free-fall distances.
2. The g cancels, so the answer does not depend on its value.

Answer: (c) at \(\dfrac{8H}{9}\) from the ground

Explanation:
1. Fall distance \(\propto t^2\)
2. At \(t/3\) the dropped distance is \(\tfrac19H\)
3. Height from the ground \(=H-\tfrac H9=\dfrac{8H}{9}\)

Teacher's Note:
1. The question asks for height above the ground, not the distance dropped.
2. Subtract the dropped part from H.

Answer: (c) \(t_2=\dfrac{t_1}{2}\)

Explanation:
1. \(h=\tfrac12gt_1^2\) and \(\tfrac h4=\tfrac12gt_2^2\)
2. Dividing, \(t_2^2=\dfrac{t_1^2}{4}\)
3. \(t_2=\dfrac{t_1}{2}\)

Teacher's Note:
1. Since \(s\propto t^2\), a quarter of the distance needs half the time.
2. Do not assume time scales linearly with distance.

Answer: (c) 7 s

Explanation:
1. Take \(g=10\): total height \(=\tfrac12\times10\times25=125\) m
2. Dropped in 3 s \(=45\) m, so remaining height \(=80\) m
3. Released from rest, \(80=5t^2\) gives \(t=4\) s
4. Total time \(=3+4=7\) s

Teacher's Note:
1. Stopping the stone resets its speed to zero, so it restarts from rest.
2. Time already used and the new fall time are added.

Answer: (c) 58.8 m

Explanation:
1. First body has been falling for 4 s: \(\tfrac12(9.8)(16)=78.4\) m
2. Second body has been falling for 2 s: \(\tfrac12(9.8)(4)=19.6\) m
3. Separation \(=78.4-19.6=58.8\) m

Teacher's Note:
1. Compute each fall distance separately from the common release point.
2. The separation grows with time because the speeds differ.

Answer: (b) 78.4 m

Explanation:
1. Height of first \(=\tfrac12(9.8)(5)^2=122.5\) m
2. Second ball falls for only 3 s: \(\tfrac12(9.8)(3)^2=44.1\) m
3. Difference \(=122.5-44.1=78.4\) m

Teacher's Note:
1. Each ball falls for a different duration but lands at the same moment.
2. Use the actual fall time of each ball in \(\tfrac12gt^2\).

Answer: (a) 35 m, 25 m and 15 m

Explanation:
1. Fall times at that instant are 4 s, 3 s, 2 s and 1 s
2. Distances dropped: 80 m, 45 m, 20 m, 5 m
3. Gaps: 35 m, 25 m, 15 m

Teacher's Note:
1. Each marble has been falling for one second less than the one before it.
2. Gaps between successive marbles form an arithmetic progression.