Q: Which graph represents uniform motion (constant velocity) in a distance–time plot?
Answer: A zig-zag line
Explanation: In a distance–time graph, slope represents speed. Constant speed means constant slope, which is a straight line. Therefore, a straight line with constant slope indicates uniform motion.
Q: A runner completes 400 m in 50 s. What is the average speed?
Answer: 7 m/s
Explanation: Average speed = distance/time. So speed = 400/50 = 8 m/s. Hence, 8 m/s is correct.
Q: Instantaneous speed is the speed?
Answer: At a particular instant of time
Explanation: Instantaneous speed is the magnitude of velocity at a specific instant. It corresponds to the limit of average speed as time interval approaches zero. A speedometer shows instantaneous speed.
Q: A wheel rotates at 120 rpm. What is its angular speed in rad/s? (Take π ≈ 3.14)?
Answer: 2π rad/s
Explanation: 120 rpm means 120/60 = 2 revolutions per second. Angular speed ω = 2×2π = 4π rad/s. Therefore, 4π rad/s is correct.
Q: In uniform circular motion, the acceleration of a particle is directed?
Answer: Opposite to velocity always
Explanation: In circular motion, acceleration is centripetal. It always points toward the center of the circle. This acceleration changes the direction of velocity, not its magnitude.
Q: A body moves with uniform acceleration from rest and reaches 12 m/s in 6 s. How much distance does it cover in this time?
Answer: 48 m
Explanation: With uniform acceleration, average velocity = (u + v)/2. Here u = 0 and v = 12, so v_avg = 6 m/s. Distance s = v_avg×t = 6×6 = 36 m.
Q: Which equation is correct for displacement in uniformly accelerated motion?
Answer: s = u + at
Explanation: For constant acceleration, displacement depends on initial velocity and time. The correct relation is s = ut + (1/2)at^2. This comes from integrating velocity with time.
Q: A body covers 100 m in 5 s starting from rest with uniform acceleration. What is the acceleration?
Answer: 4 m/s^2
Explanation: Use s = (1/2)at^2 with u = 0. So 100 = (1/2)a×25, giving a = 200/25 = 8 m/s^2. Therefore, the acceleration is 8 m/s^2.
Q: In free fall from rest (no air resistance), the distances covered in successive equal time intervals are in the ratio?
Answer: 1:3:5
Explanation: For uniformly accelerated motion from rest, distance in the nth second is proportional to (2n - 1). So the successive distances follow 1, 3, 5, 7, ... This is commonly called the odd-number rule.
Q: A projectile is thrown with speed 20 m/s at 45°. Taking g = 10 m/s^2, what is its range?
Answer: 50 m
Explanation: Range R = (u^2 sin 2θ)/g. With θ = 45°, sin 90° = 1, so R = 20^2/10 = 400/10 = 40 m. Hence, the range is 40 m.