kinematic-equations-for-constant-acceleration MCQs for UPSC Prelims

25 practice questions on kinematic-equations-for-constant-acceleration from the Describing Motion Around Us section of the UPSC Prelims syllabus. 25 come with a written explanation. Try the sample set below - the answer stays hidden until you ask for it.

8 Easy 12 Medium 5 Hard

Sample questions

Q1
medium
A stone is released from rest at the top of a tower of known height and falls freely with the uniform acceleration $g$. The time it takes to reach the ground is wanted. Which equation gives that time in one step?
  1. A $s = ut + \frac{1}{2}at^{2}$
  2. B $v^{2} = u^{2} + 2as$
  3. C $a = \frac{v - u}{s}$
  4. D $v = u + at$
Show answer and explanation

Correct answer: A - $s = ut + \frac{1}{2}at^{2}$

The known quantities are $u = 0$, $a = g$ and the height $s$, and the wanted quantity is $t$, so the equation containing all four is $s = ut + \tfrac{1}{2}at^{2}$. The equation $v = u + at$ also contains $t$ but needs the landing velocity, which is not given. The equation $v^{2} = u^{2} + 2as$ has no $t$ in it at all, and dividing a change in velocity by a distance does not define acceleration.
Q2
medium
A ball rolling down a straight slope speeds up uniformly from $3\ \text{m s}^{-1}$ to $15\ \text{m s}^{-1}$ with an acceleration of $4\ \text{m s}^{-2}$. How long does this take?
  1. A $3\ \text{s}$
  2. B $3.75\ \text{s}$
  3. C $4.5\ \text{s}$
  4. D $48\ \text{s}$
Show answer and explanation

Correct answer: A - $3\ \text{s}$

From $v = u + at$, the time is $\frac{15 - 3}{4} = \frac{12}{4} = 3\ \text{s}$. Adding the velocities instead of subtracting them gives $\frac{18}{4} = 4.5\ \text{s}$. Using only the final velocity gives $\frac{15}{4} = 3.75\ \text{s}$, which would be right only if the ball had started from rest, and $48\ \text{s}$ comes from multiplying by the acceleration instead of dividing.
Q3
hard
A stone is thrown vertically upward with a speed of $9.8\ \text{m s}^{-1}$. Taking $g = 9.8\ \text{m s}^{-2}$ and ignoring air resistance, how high above the throwing point does it rise?
  1. A $19.6\ \text{m}$
  2. B $1\ \text{m}$
  3. C $4.9\ \text{m}$
  4. D $9.8\ \text{m}$
Show answer and explanation

Correct answer: C - $4.9\ \text{m}$

At the top the velocity is zero, so $0 = u^{2} - 2gh$ gives $h = \frac{9.8^{2}}{2 \times 9.8} = 4.9\ \text{m}$. Leaving the factor $2$ out of the denominator gives $9.8\ \text{m}$. The value $1$ is $\frac{u}{g}$, which is the rise time in seconds and not a height, and $19.6\ \text{m}$ comes from doubling the starting speed instead of halving it.
Q4
medium
A motorcyclist knows his starting velocity, his uniform acceleration and the time for which he accelerated, and he wants to find the distance he covered in that time. Which equation fits the quantities he has?
  1. A $v^{2} = u^{2} + 2as$
  2. B $s = ut + \frac{1}{2}at^{2}$
  3. C $v = u + at$
  4. D $s = ut$
Show answer and explanation

Correct answer: B - $s = ut + \frac{1}{2}at^{2}$

He has $u$, $a$ and $t$ and wants $s$, and $s = ut + \tfrac{1}{2}at^{2}$ is the equation built from exactly those. The equation $v = u + at$ would give his final velocity, not a distance. The equation $v^{2} = u^{2} + 2as$ needs the final velocity, which he does not have. Writing $s = ut$ ignores the acceleration and is right only for steady motion.
Q5
hard
A car travelling at $15\ \text{m s}^{-1}$ can be brought to rest in $15\ \text{m}$ by braking uniformly. If the same car is travelling at $30\ \text{m s}^{-1}$ and brakes with exactly the same deceleration, what distance will it need to stop?
  1. A $45\ \text{m}$
  2. B $60\ \text{m}$
  3. C $15\ \text{m}$
  4. D $30\ \text{m}$
Show answer and explanation

Correct answer: B - $60\ \text{m}$

The deceleration is $\frac{15^{2}}{2 \times 15} = 7.5\ \text{m s}^{-2}$, so at $30\ \text{m s}^{-1}$ the distance is $\frac{900}{15} = 60\ \text{m}$. Because the stopping distance depends on the square of the speed, doubling the speed multiplies it by four, not by two, so $30\ \text{m}$ is wrong. The value $45\ \text{m}$ trebles the original, and $15\ \text{m}$ leaves it unchanged.
Q6
hard
A ball is thrown vertically upward from the ground with a velocity of $19.6\ \text{m s}^{-1}$. Taking upward as positive, $g = 9.8\ \text{m s}^{-2}$ and ignoring air resistance, what is the ball's velocity $3\ \text{s}$ after it was thrown?
  1. A $0\ \text{m s}^{-1}$
  2. B $9.8\ \text{m s}^{-1}$
  3. C $49\ \text{m s}^{-1}$
  4. D $-9.8\ \text{m s}^{-1}$
Show answer and explanation

Correct answer: D - $-9.8\ \text{m s}^{-1}$

Using $v = u + at$ with $a = -9.8\ \text{m s}^{-2}$ gives $19.6 - 9.8 \times 3 = -9.8\ \text{m s}^{-1}$, so after $3\ \text{s}$ the ball is falling at $9.8\ \text{m s}^{-1}$. A positive $9.8\ \text{m s}^{-1}$ has the right size but points the wrong way. Taking the acceleration as positive gives $49\ \text{m s}^{-1}$. A velocity of zero belongs to the highest point, which the ball passed at $2\ \text{s}$.
Q7
medium
A lorry travelling at $12\ \text{m s}^{-1}$ is braked uniformly and stops after covering $18\ \text{m}$. Consider these statements. (i) The magnitude of its deceleration is $4\ \text{m s}^{-2}$. (ii) The braking took $3\ \text{s}$. Which of them is correct?
  1. A Neither (i) nor (ii)
  2. B Only (i)
  3. C Only (ii)
  4. D Both (i) and (ii)
Show answer and explanation

Correct answer: D - Both (i) and (ii)

The deceleration is $\frac{u^{2}}{2s} = \frac{144}{36} = 4\ \text{m s}^{-2}$, so the first statement holds. Putting that into $v = u + at$ gives $t = \frac{12}{4} = 3\ \text{s}$, so the second holds as well. Both figures follow from the same two given values, so neither can be rejected.
Q8
medium
A car moving along a straight road at $20\ \text{m s}^{-1}$ brakes uniformly at $5\ \text{m s}^{-2}$ until it stops. What distance does it cover while braking?
  1. A $200\ \text{m}$
  2. B $4\ \text{m}$
  3. C $40\ \text{m}$
  4. D $80\ \text{m}$
Show answer and explanation

Correct answer: C - $40\ \text{m}$

Setting $v = 0$ gives $s = \frac{u^{2}}{2a} = \frac{400}{10} = 40\ \text{m}$. Leaving the factor $2$ out of the denominator gives $80\ \text{m}$. Halving $u^{2}$ without dividing by the deceleration gives $200\ \text{m}$, and $4$ is $\frac{20}{5}$, which is the braking time in seconds rather than a distance.

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