Uniform and Variable Acceleration

Uniform and Variable Acceleration MDCAT MCQs with answers

7 past paper MCQs on Uniform and Variable Acceleration, from the Force and Motion unit of MDCAT Physics. The year each question appeared is shown where known; tap “Show answer” for the correct option and a short solution.

Read the notes first Practise the unit as a quiz

1.A car starts from rest and moves with a uniform acceleration of 3m/s^2. What will be its velocity after 5 seconds? (KMU MDCAT 2025)

  1. 8m/s
  2. 12m/s
  3. 15m/s
  4. 18m/s
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Correct answer: (c) 15m/s

$v=at=3\times5$, which is $15\,\mathrm{m\,s^{-1}}$.

2.Unequal changes occurring in velocity of a body is called (SZABMU MDCAT 2025)

  1. Uniform acceleration
  2. Instantaneous acceleration
  3. variable acceleration
  4. Uniform velocity
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Correct answer: (c) variable acceleration

Unequal changes of velocity in equal times are a variable acceleration.

3.If the car is slowing down along negative x axis then acceleration will be along: (NUMS MDCAT 2023)

  1. Positive x Axis
  2. Negative x Axis
  3. Positive y Axis
  4. Negative y Axis
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Correct answer: (a) Positive x Axis

Slowing down means the acceleration points against the motion, so along the positive axis.

4.If we drop an object, it’s initial velocity is zero. How far will it fall in time ‘t’? (PMC MDCAT 2020)

  1. 9.8 t2
  2. 4.9 t2
  3. 0.49 t2
  4. 98 t2
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Correct answer: (b) 4.9 t2

$s=\tfrac12gt^2$, and half of $9.8$ is $4.9$.

5.If slope of velocity time graph is not constant at different points then body is moving with: (UHS MDCAT 2018)

  1. Uniform velocity
  2. changing acceleration
  3. Average acceleration
  4. Constant acceleration
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Correct answer: (b) changing acceleration

A changing slope means the acceleration is not constant.

6.A stone of mass 2.0 kg is dropped from a rest position 5.0 m above the ground. What is its velocity at a height of 3.0 m above the ground? (UHS MDCAT 2018)

  1. 12.5 m/s
  2. 6.3m/s
  3. 9.3 m/s
  4. 16.0 m/s
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Correct answer: (b) 6.3m/s

It has fallen $2\,\mathrm{m}$, so $v=\sqrt{2\times9.8\times2}$, about $6.3\,\mathrm{m\,s^{-1}}$.

7.A cyclist is travelling at 15 ms-1, she applies brakes so that she doesn’t collide with the wall in front of her at a distance of 18 m. Calculate the magnitude of deceleration. (UHS MDCAT 2018)

  1. 6.3 ms-1
  2. 5.3 ms-1
  3. 13 ms-1
  4. 12.5 ms-1
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Correct answer: (a) 6.3 ms-1

$v^2=2as$ gives $225=36a$, so the deceleration is about $6.3\,\mathrm{m\,s^{-2}}$.

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