Motion and Forces

20 Problems with Detailed Solutions
(Covering all five sections of “Motion and Forces”)

Section 1: Displacement & Kinematics

1. Variable-Speed Journey

   A car’s velocity is \(v(t)=4t\) (m/s) for \(0\le t\le 5\ \text{s}\).
   
   a) Displacement from \(t=0\) to \(5\ \text{s}\).
   
   b) Acceleration at \(t=2\ \text{s}\).

   Solution.
   a) \( s=\int_{0}^{5} 4t\,dt=[2t^{2}]_{0}^{5}=50~\text{m}.\)
   b) \( a=\frac{dv}{dt}=4~\text{m s}^{-2}\) (constant).
2. Change of Direction

   A cyclist travels \(100~\text{m}\) east in \(20~\text{s}\), then \(60~\text{m}\) west in \(15~\text{s}\).
   
   a) Total distance and displacement.
   
   b) Average speed and average-velocity magnitude.

   Solution.
   a) Distance \(=160~\text{m}\); Displacement \(=100-60=40~\text{m}\) east.
   b) \(\text{Avg speed}=160/35\approx 4.57~\text{m s}^{-1}\);
   \(\ |\vec v_{\text{avg}}|=40/35\approx 1.14~\text{m s}^{-1}\) east.
3. Instantaneous vs. Average

   A particle’s position is \(s(t)=t^{3}-6t^{2}+9t\) (m).
   
   a) Average velocity on \(t\in[1,3]\).
   
   b) Instantaneous velocity at \(t=2~\text{s}\).

   Solution.
   a) \(s(3)=0,\ s(1)=4 \Rightarrow \Delta s=-4,\ \Delta t=2\Rightarrow v_{\text{avg}}=-2~\text{m s}^{-1}.\)
   b) \(v(t)=3t^{2}-12t+9\Rightarrow v(2)=-3~\text{m s}^{-1}.\)
4. Non-Uniform Motion

   A runner’s speed increases uniformly from \(5\) to \(9~\text{m s}^{-1}\) in \(4~\text{s}\), then remains constant.
   
   a) Sketch \(v\)–\(t\). b) Distance in first \(4~\text{s}\) and next \(6~\text{s}\).

   Solution.
   a) Line from \((0,5)\) to \((4,9)\), then horizontal at \(v=9\).
   b) First \(4~\text{s}\): trapezoid area \(=\frac{(5+9)}{2}\cdot 4=28~\text{m}\). Next \(6~\text{s}\): \(9\cdot 6=54~\text{m}\).
5. Rapid Deceleration

   A car at \(25~\text{m s}^{-1}\) brakes with constant \(a=-5~\text{m s}^{-2}\).
   
   a) Time to stop. b) Braking distance.

   Solution.
   a) \(0=u+at\Rightarrow t=25/5=5~\text{s}\).
   b) \(s=ut+\frac12 a t^{2}=25\cdot 5+\frac12(-5)\cdot 25=62.5~\text{m}\).

Section 2: Motion Graphs

6. Piecewise \(v\)–\(t\) Analysis

   Given \(v\)–\(t\): \(0\!\to\!4~\text{s}\) at \(2\) m/s; \(4\!\to\!6~\text{s}\) linear to \(6\) m/s; \(6\!\to\!10~\text{s}\) constant at \(6\) m/s. Find total displacement.

   Solution. Areas: \(0\)–\(4\): \(2\cdot 4=8\). \(4\)–\(6\): trapezoid \(=\frac{2+6}{2}\cdot 2=8\). \(6\)–\(10\): \(6\cdot 4=24\). Total \(=40~\text{m}\).
7. Reading \(s\)–\(t\) Curves

   \(s(t)=2t^{2}\).
   
   a) \(v\) at \(t=3~\text{s}\). b) \(a\).

   Solution. \(v=\frac{ds}{dt}=4t\Rightarrow v(3)=12~\text{m s}^{-1}\). \(a=\frac{dv}{dt}=4~\text{m s}^{-2}\).
8. Acceleration–Time to Velocity

   An \(a\)–\(t\) graph is horizontal at \(a=3~\text{m s}^{-2}\) from \(0\) to \(5~\text{s}\). If \(u=2~\text{m s}^{-1}\), find \(v(t)\) and distance traveled.

   Solution. \(v(t)=u+at=2+3t\). Distance \(=\int_{0}^{5}(2+3t)\,dt=[2t+1.5t^{2}]_{0}^{5}=47.5~\text{m}\).
9. Slope of \(s\)–\(t\)

   Line through \((2,6~\text{m})\) and \((5,15~\text{m})\). What is the speed?

   Answer. \(\text{slope}=\Delta s/\Delta t=(15-6)/(5-2)=3~\text{m s}^{-1}\).
10. Curved vs. Straight

    Explain why a curved \(s\)–\(t\) graph always implies non-zero acceleration.

    Answer. Because the slope (velocity) changes with time \(\Rightarrow \frac{d^{2}s}{dt^{2}}\neq 0\).

Section 3: Uniformly Accelerated Motion

11. Uphill Skate

    A skateboarder goes up a \(5~\text{m}\) slope with \(u=6~\text{m s}^{-1}\), coming to rest halfway. Find \(a\).

    Solution. Over \(s=2.5~\text{m}\): \(0=u^{2}+2as\Rightarrow a=-\frac{36}{5}=-7.2~\text{m s}^{-2}\).
12. Downward Throw

    A stone is thrown downward at \(5~\text{m s}^{-1}\) from \(20~\text{m}\). Time to hit ground?

    Solution. \(s=ut+\frac12 gt^{2}\Rightarrow 20=5t+4.9t^{2}\Rightarrow t\approx 1.57~\text{s}\).
13. Upward Launch

    A football is kicked at \(15~\text{m s}^{-1}\) upward.
    
    a) Time to max height. b) Max height.

    Solution. a) \(t=\frac{u}{g}=\frac{15}{9.8}\approx 1.53~\text{s}\).
    b) \(s_{\max}=\frac{u^{2}}{2g}=\frac{225}{19.6}\approx 11.48~\text{m}\).
14. Braking Distance

    A car at \(20~\text{m s}^{-1}\) decelerates at \(4~\text{m s}^{-2}\). How far to stop?

    Solution. \(0=u^{2}+2as\Rightarrow s=\frac{u^{2}}{2|a|}=\frac{400}{8}=50~\text{m}\).
15. Mid-race Speed

    From rest, \(a=3~\text{m s}^{-2}\) for \(8~\text{s}\), then coast. Speed at \(t=8~\text{s}\)?

    Answer. \(v=at=3\cdot 8=24~\text{m s}^{-1}\).

Section 4: Linear Momentum

16. Crash Test

    A \(1000~\text{kg}\) car at \(20~\text{m s}^{-1}\) stops in \(0.1~\text{s}\). Average force on barrier?

    Solution. \(\Delta p=1000\cdot 20=2.0\times 10^{4}\). \(F_{\text{avg}}=\Delta p/\Delta t=2.0\times 10^{5}~\text{N}\).
17. Recoil of Gun

    A \(0.01~\text{kg}\) bullet leaves a \(5~\text{kg}\) rifle at \(400~\text{m s}^{-1}\). Recoil speed?

    Solution. \(0=m_R v_R+m_b v_b\Rightarrow v_R=-(0.01\cdot 400)/5=-0.8~\text{m s}^{-1}\).
18. Two-Mass Collision

    \(m_1=2~\text{kg}\) at \(u_1=3~\text{m s}^{-1}\) hits \(m_2=1~\text{kg}\) at rest, elastically. Speeds after?

    Solution.
    \(v_1’=\frac{m_1-m_2}{m_1+m_2}u_1=\frac{1}{3}\cdot 3=1~\text{m s}^{-1}\).
    \(v_2’=\frac{2m_1}{m_1+m_2}u_1=\frac{4}{3}\cdot 3=4~\text{m s}^{-1}\).
19. Inelastic Impact

    A \(0.2~\text{kg}\) ball at \(5~\text{m s}^{-1}\) sticks to a \(0.8~\text{kg}\) block at rest. Final velocity?

    Solution. \(v=\frac{0.2\cdot 5 + 0}{1.0}=1.0~\text{m s}^{-1}\).
20. Rocket Propulsion (Simplified)

    A rocket ejects \(100~\text{kg}\) of gas at \(2000~\text{m s}^{-1}\). Initial mass \(1000~\text{kg}\). \(\Delta v\)?

    Solution (Tsiolkovsky). \(\Delta v=v_e\ln\!\left(\frac{m_0}{m_1}\right)=2000\ln(1000/900)\approx 210.6~\text{m s}^{-1}\).

Section 5: Newton’s Laws

21. Atwood Machine

    \(m_1=5~\text{kg},\ m_2=3~\text{kg}\) over a frictionless pulley. Find \(a\) and \(T\).

    Solution. \(a=\frac{(m_1-m_2)g}{m_1+m_2}=\frac{2\cdot 9.8}{8}=2.45~\text{m s}^{-2}\).
    \(T=m_1(g-a)=5(9.8-2.45)=\mathbf{36.75~\text{N}}\).
22. Block on Incline with Friction

    \(m=4~\text{kg},\ \theta=30^\circ,\ \mu_k=0.20\). Find \(a\) down the slope.

    Solution.
    \(F_{\parallel}=mg\sin\theta=4\cdot 9.8\cdot 0.5=19.6~\text{N}\).
    \(f_k=\mu_k N=\mu_k mg\cos\theta=0.20\cdot 4\cdot 9.8\cdot 0.866\approx 6.79~\text{N}\).
    Net \(=19.6-6.79=12.81~\text{N}\Rightarrow a=12.81/4\approx \mathbf{3.20~\text{m s}^{-2}}\).
23. Centripetal Force

    A \(0.5~\text{kg}\) mass on a \(2~\text{m}\) string circles at \(2~\text{rev s}^{-1}\). Tension?

    Solution. \(\omega=2(2\pi)=4\pi\ \text{rad s}^{-1}\).
    \(F=m\omega^{2}r=0.5(4\pi)^{2}\cdot 2=16\pi^{2}\approx 158~\text{N}\).
24. Elevator Tension

    Elevator \(m=600~\text{kg}\) accelerates upward at \(1.5~\text{m s}^{-2}\). Cable tension?

    Solution. \(T-mg=ma\Rightarrow T=m(g+a)=600(9.8+1.5)=\mathbf{6780~\text{N}}\).
25. Forced Oscillation (Hooke + Newton)

    Mass \(m=0.5~\text{kg}\) on spring \(k=100~\text{N m}^{-1}\) pulled to \(x=0.1~\text{m}\) and released. Max acceleration?

    Solution. \(a_{\max}=\frac{k}{m}x=\frac{100}{0.5}\cdot 0.1=20~\text{m s}^{-2}\).