Motion in a Straight Line — Class 11 MCQs with Answers

Motion in a Straight Line — important questions & MCQs with answers (Class 11 Physics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Velocity is:

   AScalarBMass × timeCSame as speedDVector
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   Answer: Vector

   Explanation: Velocity is a vector: it has both magnitude and direction, e.g. 5 m/s due north, unlike speed (5 m/s alone). Calling it 'same as speed' drops the direction — that is the trap.

2. Q2 of 6Easy

   SI unit of acceleration:

   Am/sBm/s²CmDkg·m/s²
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   Answer: m/s²

   Explanation: Acceleration a = Δv/Δt, so its unit is (m/s)/s = m/s², e.g. gaining 10 m/s every second is 10 m/s². Confusing it with velocity's unit m/s is the common slip.

3. Q3 of 6Medium

   Object falls 5 m in 1st second. Distance in 2nd second (g = 10):

   A10 mB5 mC15 mD20 m
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   Answer: 15 m

   Explanation: With g = 10 m/s², total fall by t = 2 s is ½×10×4 = 20 m, and by t = 1 s it is ½×10×1 = 5 m, matching the given 1st-second value. Distance in the 2nd second = 20 − 5 = 15 m, not simply double the 5 m.

4. Q4 of 6Medium

   Body's x-t graph is a straight line (non-horizontal). Body has:

   

   AConstant velocityBConstant accelerationCVariable velocityDNo motion
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   Answer: Constant velocity

   Explanation: A straight, non-horizontal x-t line has constant slope, and since slope = dx/dt = velocity, that means constant velocity, e.g. rising 4 m every 2 s gives a steady 2 m/s. A curved x-t graph signals acceleration, not a straight one.

5. Q5 of 6Hard

   Slope of position-time graph gives:

   AMassBAccelerationCForceDVelocity
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   Answer: Velocity

   Explanation: Instantaneous velocity is v = dx/dt, so the slope of a position-time graph directly gives velocity, e.g. a slope of 3 m/s means 3 m displacement each second — acceleration instead needs a v-t graph's slope.

6. Q6 of 6Hard

   Body decelerates from 20 m/s to rest in 4 s. Magnitude of acceleration:

   A20 m/s²B4 m/s²C5 m/s²D80 m/s²
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   Answer: 5 m/s²

   Explanation: Deceleration magnitude |a| = |v − u|/t = |0 − 20|/4 = 5 m/s². Dividing 20 by 4 correctly gives 5, not 4 — don't confuse the given time with the answer.

Motion in a Straight Line notes: key concepts (Class 11 Physics)

One-dimensional kinematics: the difference between distance and displacement, speed and velocity, how position-time and velocity-time graphs encode motion, the three equations of uniformly accelerated motion, free fall under gravity, and relative velocity along a line.

Distance versus displacement

Distance is the total path length, a scalar that never decreases. Displacement is the straight-line change in position, a vector that can be zero even after a long journey.

Average speed versus average velocity

Average speed uses path length, so it can never be less than the magnitude of average velocity. The two differ whenever the motion reverses direction.

Instantaneous velocity

The limit of average velocity as the time interval shrinks to zero — mathematically dx/dt, and graphically the slope of the tangent to the position-time curve.

Acceleration

The rate of change of velocity. A negative acceleration does not mean slowing down; a body slows only when its velocity and acceleration point in opposite directions.

Position-time graph

Its slope at any instant gives velocity. A straight line means uniform velocity; a curve means the velocity is changing, so acceleration is non-zero.

Velocity-time graph

Its slope gives acceleration and the AREA under it gives displacement, with area below the time axis counting as negative displacement.

Equations of motion

The three kinematic equations hold only for motion in a straight line with CONSTANT acceleration. They break down the instant acceleration varies.

Displacement in the nth second

The displacement during one particular second of the motion, not the total over n seconds. It grows linearly with n for constant acceleration.

Free fall

Motion under gravity alone, with air resistance neglected. The acceleration is g ≈ 9.8 m s⁻² downward and is completely independent of the body's mass.

Vertical throw

At the highest point the velocity is zero but the acceleration is still g downward. For the same level, time of ascent equals time of descent.

Relative velocity

The velocity of A as measured by B, equal to v_A − v_B. For bodies approaching head-on the speeds add; for bodies moving the same way they subtract.

Uniform versus non-uniform motion

Uniform motion covers equal displacements in equal intervals, so acceleration is zero. Non-uniform motion has changing velocity and therefore non-zero acceleration.

Stopping distance

The distance a braking body travels before stopping. Because it depends on the SQUARE of the initial speed, doubling the speed quadruples the stopping distance.