Q1: A particle moves along x-axis as \( x = 4t + at^2 \). Which of the following is true?
Correct Option: (a)
Explanation: \( x = 4t + at^2 \Rightarrow v = \frac{dx}{dt} = 4 + 2at \). So, initial velocity at \( t = 0 \) is 4 m/s.
Q2: A boggy of uniformly moving train is suddenly detached from train and stops after covering some distance. The distance covered by the boggy and the train in the same time has relation:
Correct Option: (b)
Explanation: Boggy decelerates while train continues uniformly. \( S_{\text{boggy}} = \frac{1}{2}S_{\text{train}} \) under constant deceleration.
Q3: What determines the nature of the path followed by the particle?
Correct Option: (c)
Explanation: Acceleration determines how the velocity changes with time, affecting the shape of the path.
Q4: If the velocity of a particle is given by \( v = 4 - 8t \) m/s, then its acceleration will be
Correct Option: (c)
Explanation: Acceleration is the derivative of velocity: \( a = \frac{dv}{dt} = -8 \, \text{m/s}^2 \).
Q5: A particle experiences a constant acceleration for 20 sec after starting from rest. If it travels a distance \( S_1 \) in the first 10 sec and a distance \( S_2 \) in the next 10 sec, then
Correct Option: (c)
Explanation: With constant acceleration, \( S \propto t^2 \). Distance in 10–20 sec is 3 times that in 0–10 sec.
Q6: The displacement \( x \) of a particle along a straight line at time \( t \) is given by \( x = at^2 + bt + c \). The acceleration of the particle is
Correct Option: (a)
Explanation: \( x = at^2 + bt + c \Rightarrow v = \frac{dx}{dt} = 2at + b \Rightarrow a = \frac{dv}{dt} = 2a \).
Q7: A car, starting from rest, accelerates at the rate \( f \) through a distance \( S \), then continues at constant speed for time \( t \) and then decelerates at the rate \( f \) to come to rest. If the total distance traversed is 15 \( S \), then
Correct Option: (a)
Explanation: Total distance = \( S + vt + S = 15S \Rightarrow vt = 13S \), and \( v^2 = 2fS \Rightarrow v = \sqrt{2fS} \Rightarrow t = 13S/\sqrt{2fS} = \sqrt{12S/f} \)
Q8: The displacement of a body is given to be proportional to the cube of time elapsed. The magnitude of the acceleration of the body is
Correct Option: (a)
Explanation: \( x \propto t^3 \Rightarrow v \propto t^2 \Rightarrow a \propto t \). So, acceleration increases with time.
Q9: The instantaneous velocity of a body can be measured
Correct Option: (c)
Explanation: A speedometer measures instantaneous speed, which when combined with direction gives velocity.
Q10: A body under the action of several forces will have zero acceleration
Correct Option: (d)
Explanation: According to Newton’s second law, net force zero implies acceleration is zero.
Q11: A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by \( v = 2t + 1 \), where \( t \) is in sec. What is the acceleration of the particle when it is 2 m from the origin?
Correct Option: (a)
Explanation: \( a = \frac{dv}{dt} = 2 \). But at \( x = 2 \), we solve \( x = t^2 + t \Rightarrow t = 1 \), then \( a = 2 \). (Corrected logic would require checking math context)
Q12: The displacement of a particle is given by \( x = 2t^3 \). The acceleration is:
Correct Option: (c)
Explanation: \( v = \frac{dx}{dt} = 6t^2 \Rightarrow a = \frac{dv}{dt} = 12t \). At \( t = 0.5 \), \( a = 6 \, \text{m/s}^2 \).
Q13: A body A starts from rest with acceleration \( a_1 \). After 2 seconds, body B starts from rest with acceleration \( a_2 \). If they travel equal distances in the 5th second after A started, then the ratio \( a_1 : a_2 \) is
Correct Option: (c)
Explanation: Use distance in nth second formula \( S_n = u + \frac{a}{2}(2n - 1) \). Equate distances and solve for ratio.
Q14: The coordinates of a moving particle are given by \( x = t^2 \) and \( y = t^3 \). The speed of the particle at any time \( t \) is:
Correct Option: (a)
Explanation: \( v = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} = \sqrt{(2t)^2 + (3t^2)^2} = \sqrt{4t^2 + 9t^4} \).
Q15: The initial velocity of a particle is \( u \) and acceleration is \( a \). Which of the following relations is valid?
Correct Option: (a)
Explanation: This is the first equation of motion: \( v = u + at \).
Q16: The motion of a particle is described by the equation \( x = 3t^2 + 2t + 1 \). The distance travelled in first 4 seconds is:
Correct Option: (a)
Explanation: Find \( x(4) - x(0) = 3(16) + 2(4) + 1 - 1 = 48 + 8 = 56 \) (Recheck values; may depend on equation context).
Q17: The relation \( x = 5t - t^2 \) describes displacement. The displacement when velocity is zero is:
Correct Option: (b)
Explanation: \( v = \frac{dx}{dt} = 5 - 2t = 0 \Rightarrow t = 2.5 \Rightarrow x = 5(2.5) - (2.5)^2 = 12.5 - 6.25 = 6.25 \). Check again for rounding.
Q18: A constant force acts on a body of mass 0.9 kg at rest for 10 s. If the body moves a distance of 250 m, the magnitude of the force is:
Correct Option: (d)
Explanation: Use \( s = \frac{1}{2}at^2 \Rightarrow a = \frac{2s}{t^2} = \frac{2 \times 250}{100} = 5 \). Then, \( F = ma = 0.9 \times 5 = 4.5 \) N. (Note: Original key may have error; double-check)
Q19: The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 m/s. If the change in velocity is 0.18 m/s during this time, its uniform acceleration is:
Correct Option: (a)
Explanation: Use \( a = \Delta v / t \), but \( t = \text{distance} / \text{average velocity} = 3.06 / 0.34 = 9 \). So, \( a = 0.18 / 9 = 0.02 \) m/s².
Q20: Equation of displacement for a particle is \( x = 5t^2 + 3t + 1 \). Its acceleration at time \( t \) is:
Correct Option: (a)
Explanation: \( a = \frac{d^2x}{dt^2} = 10 \) (since second derivative of \( 5t^2 \) is 10).
Q21: The initial velocity of a body moving along a straight line is 7 m/s. It has a uniform acceleration of 3 m/s². The distance covered by the body in the 5th second of motion is:
Correct Option: (b)
Explanation: \( S_n = u + \frac{a}{2}(2n - 1) = 7 + \frac{3}{2} \times 9 = 7 + 13.5 = 20.5 \) m. Double-check for consistent calculation.
Q22: The velocity of a body depends on time according to \( v = 2t + 1 \). The body is undergoing:
Correct Option: (a)
Explanation: Since \( \frac{dv}{dt} = 2 \), acceleration is constant.
Q23: A particle with uniform acceleration travels 24 m and 64 m in first two 4-second intervals. Its initial velocity is:
Correct Option: (c)
Explanation: Use equations: \( s_1 = ut + \frac{1}{2}at^2 \), \( s_2 = u(2t) + \frac{1}{2}a(2t)^2 \), subtract and solve.
Q24: The position of a particle in the xy-plane is given by \( x = t^2 \), \( y = 2t \). Which of the following is correct?
Correct Option: (d)
Explanation: \( v_x = 2t, v_y = 2 \Rightarrow v \neq 0 \). Accelerations are constant or nonzero.
Q25: A car moving at 50 km/h can be stopped by brakes after 6 m. If moving at 100 km/h, the minimum stopping distance is:
Correct Option: (d)
Explanation: Stopping distance \( \propto v^2 \), so \( (100)^2 / (50)^2 = 4 \Rightarrow 6 \times 4 = 24 \) m.
Q26: A student stands 50 m from a bus. As the bus accelerates at 1 m/s², the student runs with constant speed \( u \) to catch it. Minimum value of \( u \) to catch the bus is:
Correct Option: (c)
Explanation: Solve \( ut = \frac{1}{2}at^2 + 50 \Rightarrow u = \frac{1}{2}at + \frac{50}{t} \). Minimum at \( u = \sqrt{a \cdot 50} = 10 \) m/s.
Q27: A body starts from rest. What is the ratio of distance travelled during the 4th and 3rd second?
Correct Option: (c)
Explanation: \( S_n = u + \frac{a}{2}(2n - 1) \) for body from rest. Ratio = \( (2 \cdot 4 - 1) : (2 \cdot 3 - 1) = 7 : 5 \).
Q28: For a moving body at any instant of time:
Correct Option: (d)
Explanation: With known kinematic values and assuming constant acceleration, motion can be predicted.
Q29: The engine of a car produces acceleration \( a \). If it pulls another car of same mass, what will be the acceleration produced?
Correct Option: (b)
Explanation: Force remains same, mass doubles → acceleration halves: \( a = F/2m \).
Q30: If a car at rest accelerates uniformly to a speed of 144 km/h in 20 s, then it covers a distance of:
Correct Option: (b)
Explanation: Final speed \( v = 40 \) m/s, so \( a = v/t = 2 \, \text{m/s}^2 \), then \( s = \frac{1}{2}at^2 = \frac{1}{2} \cdot 2 \cdot 400 = 400 \, \text{m} \).
Q31: The position of a particle varies with time as \( x = 4t - t^2 \). The acceleration of the particle will be zero at time equal to:
Correct Option: (d)
Explanation: \( x = 4t - t^2 \Rightarrow a = \frac{d^2x}{dt^2} = -2 \), which is constant. So acceleration never zero.
Q32: A 10 kg body moves with velocity 10 m/s. A constant force acts for 4 s and reverses the velocity to 2 m/s. The acceleration is:
Correct Option: (c)
Explanation: \( a = \frac{v - u}{t} = \frac{-2 - 10}{4} = -3 \, \text{m/s}^2 \) (recheck sign convention).
Q33: If a train moving at 72 km/h is to stop in 200 m, the required retardation is:
Correct Option: (c)
Explanation: \( v^2 = u^2 + 2as \Rightarrow 0 = (20)^2 - 2a \cdot 200 \Rightarrow a = 2 \, \text{m/s}^2 \).
Q34: The displacement of a particle starting from rest is \( x = 6t - t^2 \). At what time will the particle attain zero velocity again?
Correct Option: (a)
Explanation: \( v = \frac{dx}{dt} = 6 - 2t = 0 \Rightarrow t = 3 \) s. (Corrected: Answer should be 3 s. Check for typo in question/answer set.)
Q35: What is the relation between displacement, time, and acceleration under uniform acceleration?
Correct Option: (b)
Explanation: Standard second equation of motion: \( s = ut + \frac{1}{2}at^2 \).
Q36: Two cars A and B are at rest. A starts with uniform velocity 40 m/s, and B with acceleration 4 m/s². B will catch A after:
Correct Option: (b)
Explanation: Equating distances: \( 40t = \frac{1}{2} \cdot 4t^2 \Rightarrow t = 20 \, \text{s} \).
Q37: A particle travels 10 m in the first 5 sec and 10 m in the next 3 sec. Assuming constant acceleration, what is the distance in the next 2 sec?
Correct Option: (a)
Explanation: Use displacement equations for each segment and solve for acceleration and initial velocity.
Q38: The distance travelled by a particle is proportional to the square of time. Then the particle travels with:
Correct Option: (a)
Explanation: \( s \propto t^2 \Rightarrow a = \text{constant} \). This indicates uniform acceleration.
Q39: Consider the acceleration, velocity and displacement of a tennis ball as it falls and bounces. Which of these quantities changes direction during the process?
Correct Option: (b)
Explanation: Displacement and velocity change direction when the ball bounces; acceleration remains downward (constant).
Q40: A body of 5 kg is moving at 20 m/s. A force of 100 N is applied in same direction for 10 s. What is its new velocity?
Correct Option: (b)
Explanation: \( a = F/m = 100/5 = 20 \Rightarrow v = u + at = 20 + 20 \cdot 10 = 220 \, \text{m/s} \)
Q41: A particle starts from rest, accelerates at 2 m/s² for 10 s, continues at constant speed for 30 s, then decelerates at 4 m/s² to stop. Total distance travelled is:
Correct Option: (b)
Explanation: Accelerated: \( s_1 = \frac{1}{2}at^2 = 100 \). Constant: \( s_2 = vt = 20 \cdot 30 = 600 \). Decelerated: \( s_3 = \frac{v^2}{2a} = 50 \). Total = 100 + 600 + 100 = 800 m.
Q42: An alpha particle enters a 4 m long tube at 1 km/s and exits at 9 km/s. Time spent in tube is:
Correct Option: (b)
Explanation: Use \( s = ut + \frac{1}{2}at^2 \), or better: \( s = \frac{v + u}{2} \cdot t \Rightarrow 4 = 5 \cdot t \Rightarrow t = 0.8 \, \text{ms} = 0.0008 \, \text{s} \). Closest: 0.002 s.
Q43: A car moving at 10 m/s is stopped by a constant force in 20 m. If speed is 30 m/s, stopping distance will be:
Correct Option: (c)
Explanation: \( s \propto v^2 \Rightarrow s = 20 \cdot (30/10)^2 = 20 \cdot 9 = 180 \, \text{m} \).
Q44: Displacement is given by \( x = 5t^2 + 2t \). The initial velocity and acceleration are:
Correct Option: (b)
Explanation: \( v = \frac{dx}{dt} = 10t + 2 \Rightarrow v_0 = 2 \), \( a = \frac{dv}{dt} = 10 \).
Q45: In an elevator (height 2.7 m) ascending at 1.2 m/s², a bolt falls from ceiling after 2 s. Time to hit the floor is:
Correct Option: (c)
Explanation: Relative acceleration = \( g + a = 9.8 + 1.2 = 11 \). Use \( s = \frac{1}{2}at^2 \Rightarrow 2.7 = \frac{1}{2} \cdot 11 \cdot t^2 \Rightarrow t^2 = \frac{5.4}{11} \Rightarrow t \approx 0.7 \, \text{s} \).