Equations of Motion — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Equations of Motion MCQs with step-by-step solutions (21 questions). Part of Motion in a Plane (Std 11). Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Equations of Motion · easy · theory
The equation $v = u + at$ is valid only when the:
A. Velocity is uniform
B. Acceleration is uniform ✓ Correct
C. Body moves in a circle
D. Body starts from rest
Solution: All three kinematic equations assume a constant acceleration.
Q2 — Equations of Motion · easy · theory
The equation giving displacement in terms of time is:
A. $s = ut + at^2$
B. $s = \dfrac{1}{2}at$
C. $s = ut + \dfrac{1}{2}at^2$ ✓ Correct
D. $s = u + \dfrac{1}{2}at^2$
Solution: The first term is the displacement at constant velocity, the second the extra due to acceleration.
Q3 — Equations of Motion · easy · theory
The equation connecting velocity and displacement without time is:
A. $v^2 = u^2 + at$
B. $s = ut + \dfrac{1}{2}at^2$
C. $v = u + at$
D. $v^2 = u^2 + 2as$ ✓ Correct
Solution: It is obtained by eliminating $t$ between the other two equations.
Q4 — Equations of Motion · hard · theory
The distance covered in the $n^{\text{th}}$ second of uniformly accelerated motion is:
A. $u + an$
B. $un + \dfrac{a}{2}n^2$
C. $u + \dfrac{a}{2}(2n + 1)$
D. $u + \dfrac{a}{2}(2n - 1)$ ✓ Correct
Solution: It is the difference between the displacements in $n$ and $(n-1)$ seconds.
Q5 — Equations of Motion · easy · theory
For a body in free fall near the Earth, the acceleration is:
A. Zero
B. Dependent on the mass of the body
C. Equal to $g$ and directed vertically downward ✓ Correct
D. Equal to $g$ and directed upward
Solution: In the absence of air resistance all bodies fall with the same acceleration.
Q6 — Equations of Motion · medium · theory
When a body is thrown vertically upward, at the highest point its:
A. Acceleration is directed upward
B. Acceleration is zero but its velocity is not
C. Velocity is zero but its acceleration is still $g$ downward ✓ Correct
D. Velocity and acceleration are both zero
Solution: Gravity continues to act throughout, which is what brings the body back down.
Q7 — Equations of Motion · medium · theory
In applying the equations of motion, a consistent sign convention is needed because:
A. Displacement, velocity and acceleration are vectors ✓ Correct
B. Acceleration is always positive
C. Time can be negative
D. The equations are only approximate
Solution: Quantities directed opposite to the chosen positive direction must be entered as negative.
Q8 — Equations of Motion · hard · theory
The equations of motion can be derived from a velocity-time graph because:
A. The graph passes through the origin
B. Velocity is always constant
C. The graph is always a parabola
D. The area under the graph gives displacement and its slope gives acceleration ✓ Correct
Solution: For uniform acceleration the graph is a straight line, and the area is a simple trapezium.
Q9 — Equations of Motion · medium · numerical
A body starts from rest with an acceleration of $2\text{ m/s}^2$. Its velocity and displacement after $5\text{ s}$ are:
A. $5\text{ m/s}$ and $25\text{ m}$
B. $10\text{ m/s}$ and $50\text{ m}$
C. $10\text{ m/s}$ and $25\text{ m}$ ✓ Correct
D. $20\text{ m/s}$ and $50\text{ m}$
Solution: $v = 0 + 2(5) = 10\text{ m/s}$ and $s = 0 + \dfrac{1}{2}(2)(25) = 25\text{ m}$.
Q10 — Equations of Motion · easy · numerical
A body moving at $10\text{ m/s}$ accelerates at $2\text{ m/s}^2$ for $3\text{ s}$. Its final velocity is:
A. $13\text{ m/s}$
B. $6\text{ m/s}$
C. $16\text{ m/s}$ ✓ Correct
D. $30\text{ m/s}$
Solution: $v = u + at = 10 + 2(3) = 16\text{ m/s}$.
Q11 — Equations of Motion · medium · numerical
A body falls freely from rest. The distance it covers in $3\text{ s}$ is ($g = 10\text{ m/s}^2$):
A. $45\text{ m}$ ✓ Correct
B. $90\text{ m}$
C. $30\text{ m}$
D. $15\text{ m}$
Solution: $s = \dfrac{1}{2}gt^2 = \dfrac{1}{2}(10)(9) = 45\text{ m}$.
Q12 — Equations of Motion · medium · numerical
A body starts from rest with acceleration $2\text{ m/s}^2$ and travels $25\text{ m}$. Its final velocity is:
A. $10\text{ m/s}$ ✓ Correct
B. $100\text{ m/s}$
C. $50\text{ m/s}$
D. $5\text{ m/s}$
Solution: $v^2 = 0 + 2(2)(25) = 100$, so $v = 10\text{ m/s}$.
Q13 — Equations of Motion · medium · numerical
A stone is dropped from a height of $80\text{ m}$. The time it takes to reach the ground is ($g = 10\text{ m/s}^2$):
A. $4\text{ s}$ ✓ Correct
B. $2\text{ s}$
C. $16\text{ s}$
D. $8\text{ s}$
Solution: $80 = \dfrac{1}{2}(10)t^2$, so $t^2 = 16$ and $t = 4\text{ s}$.
Q14 — Equations of Motion · medium · numerical
A stone is dropped from a height of $45\text{ m}$. Its speed on reaching the ground is ($g = 10\text{ m/s}^2$):
A. $45\text{ m/s}$
B. $90\text{ m/s}$
C. $30\text{ m/s}$ ✓ Correct
D. $15\text{ m/s}$
Solution: $v = \sqrt{2gh} = \sqrt{2 \times 10 \times 45} = \sqrt{900} = 30\text{ m/s}$.
Q15 — Equations of Motion · hard · numerical
A body falls freely from rest. The distance it covers during the third second is ($g = 10\text{ m/s}^2$):
A. $20\text{ m}$
B. $30\text{ m}$
C. $25\text{ m}$ ✓ Correct
D. $45\text{ m}$
Solution: $s_n = \dfrac{g}{2}(2n - 1) = 5(2 \times 3 - 1) = 25\text{ m}$.
Q16 — Equations of Motion · medium · numerical
A body is thrown vertically upward at $20\text{ m/s}$. The maximum height it reaches is ($g = 10\text{ m/s}^2$):
A. $20\text{ m}$ ✓ Correct
B. $10\text{ m}$
C. $40\text{ m}$
D. $2\text{ m}$
Solution: $H = \dfrac{u^2}{2g} = \dfrac{400}{20} = 20\text{ m}$.
Q17 — Equations of Motion · easy · numerical
A body is thrown vertically upward at $20\text{ m/s}$. The time it takes to reach the highest point is ($g = 10\text{ m/s}^2$):
A. $4\text{ s}$
B. $1\text{ s}$
C. $20\text{ s}$
D. $2\text{ s}$ ✓ Correct
Solution: $t = \dfrac{u}{g} = \dfrac{20}{10} = 2\text{ s}$.
Q18 — Equations of Motion · hard · numerical
A car travelling at $20\text{ m/s}$ is brought to rest in a distance of $50\text{ m}$. Its retardation is:
A. $2\text{ m/s}^2$
B. $4\text{ m/s}^2$ ✓ Correct
C. $8\text{ m/s}^2$
D. $0.4\text{ m/s}^2$
Solution: $0 = 400 + 2a(50)$, so $a = -4\text{ m/s}^2$.
Q19 — Equations of Motion · hard · numerical
A stone dropped from rest covers $25\text{ m}$ during the last second of its fall. The height from which it was dropped is ($g = 10\text{ m/s}^2$):
A. $30\text{ m}$
B. $45\text{ m}$ ✓ Correct
C. $80\text{ m}$
D. $25\text{ m}$
Solution: $s_n = 5(2n - 1) = 25$ gives $n = 3\text{ s}$, so $h = \dfrac{1}{2}(10)(9) = 45\text{ m}$.
Q20 — Equations of Motion · hard · numerical
A body speeds up from $5\text{ m/s}$ to $15\text{ m/s}$ over a distance of $40\text{ m}$. Its acceleration is:
A. $10\text{ m/s}^2$
B. $1.25\text{ m/s}^2$
C. $2.5\text{ m/s}^2$ ✓ Correct
D. $5\text{ m/s}^2$
Solution: $a = \dfrac{v^2 - u^2}{2s} = \dfrac{225 - 25}{80} = 2.5\text{ m/s}^2$.
Q21 — Equations of Motion · easy · theory
Two bodies of different masses are dropped from the same height in the absence of air resistance. They reach the ground:
A. The heavier one first
B. At times proportional to their masses
C. At the same time, since the acceleration is independent of mass ✓ Correct
D. The lighter one first
Solution: The equation $h = \dfrac{1}{2}gt^2$ contains no mass term.