Kinematics

Practise Kinematics for FBISE Grade 9–10 Physics (Grade 9–10): 5 multiple-choice questions with instant checking and worked explanations. Free, no sign-up.

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1. Question 1

   A vehicle starts from rest, accelerates uniformly for $4\mathrm{s}$, then moves at constant speed for $6\mathrm{s}$, and finally decelerates uniformly to rest in $3\mathrm{s}$. The total distance covered is $228\mathrm{m}$. What is the magnitude of the acceleration during the first phase?
   • A$4\mathrm{m}{\mathrm{s}}^{-2}$
   • B$7\mathrm{m}{\mathrm{s}}^{-2}$
   • C$6\mathrm{m}{\mathrm{s}}^{-2}$
   • D$5\mathrm{m}{\mathrm{s}}^{-2}$
   Show answer and explanation

   Answer: C) $6\mathrm{m}{\mathrm{s}}^{-2}$

   Using the equations for uniformly accelerated motion, the distance covered in each phase is expressed in terms of the unknown acceleration a. Summing the three distances and setting the total equal to 228 m gives 38a = 228, so a = 6 m·s⁻². The other options arise from common algebraic mis‑applications of the motion formulas.

2. Question 2

   A cyclist speeds up from $4\mathrm{m}{\mathrm{s}}^{-1}$ to $8\mathrm{m}{\mathrm{s}}^{-1}$ over a distance of $40\mathrm{m}$. What is his average acceleration?
   • A$0.5\mathrm{m}{\mathrm{s}}^{-2}$
   • B$0.6\mathrm{m}{\mathrm{s}}^{-2}$
   • C$0.7\mathrm{m}{\mathrm{s}}^{-2}$
   • D$0.4\mathrm{m}{\mathrm{s}}^{-2}$
   Show answer and explanation

   Answer: B) $0.6\mathrm{m}{\mathrm{s}}^{-2}$

   The kinematic relation $v_f^2=v_i^2+2as$ directly links the change in speed to the acceleration over a known distance. Substituting the given speeds and distance yields $a=0.6\mathrm{m}{\mathrm{s}}^{-2}$. The other options arise from using the wrong formula or an incorrect time estimate.

3. Question 3

   Which velocity–time graph represents motion with constant acceleration starting from rest?
   • AA straight line with negative slope
   • BA horizontal line at positive velocity
   • CA straight line through the origin with positive slope
   • DA parabola opening upwards
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   Answer: C) A straight line through the origin with positive slope

   Constant acceleration from rest gives $v=at$, a straight line through the origin with a positive slope on a $v$–$t$ graph. The other graphs either show constant velocity, a position‑time shape, or decreasing velocity, none of which represent constant positive acceleration from rest.

4. Question 4

   What is the SI unit of acceleration?
   • AN
   • Bm/s
   • Ckm/h
   • Dm/s^2
   Show answer and explanation

   Answer: D) m/s^2

   Acceleration measures the rate of change of velocity, so its unit is metres per second squared ($\mathrm{m}{\mathrm{s}}^{-2}$). The other options are units of speed ($\mathrm{m}{\mathrm{s}}^{-1}$, km/h) or force (N), which are not appropriate for acceleration.

5. Question 5

   A cyclist starts from rest, accelerates uniformly for $5\mathrm{s}$, then rides at the attained constant speed for $8\mathrm{s}$. A car that is $120\mathrm{m}$ ahead of the cyclist moves at a constant speed of $12\mathrm{m}{\mathrm{s}}^{-1}$ and overtakes the cyclist exactly when the cyclist finishes the $8\mathrm{s}$ segment. What is the cyclist's acceleration during the first $5\mathrm{s}$?
   • A$4.8\mathrm{m}{\mathrm{s}}^{-2}$
   • B$5.0\mathrm{m}{\mathrm{s}}^{-2}$
   • C$6.0\mathrm{m}{\mathrm{s}}^{-2}$
   • D$5.3\mathrm{m}{\mathrm{s}}^{-2}$
   Show answer and explanation

   Answer: D) $5.3\mathrm{m}{\mathrm{s}}^{-2}$

   The car and cyclist meet after 13 s, so their travelled distances from the cyclist’s start must be equal. The car’s distance is $120+12\times 13=276$m. The cyclist covers $\frac{1}{2}a(5{)}^2$ during the first 5 s and then moves at the speed $v=a\times 5$ for the next 8 s, giving a total distance $12.5a+40a=52.5a$. Equating $52.5a$ to 276 m yields $a\approx 5.3{\text{m s}}^{-2}$, the correct answer.

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