Question Statement

A motorcyclist $M$ leaves a road junction at time $t=0\mathrm{s}$. He accelerates from rest at a rate of $3\mathrm{m}/{\mathrm{s}}^2$ for $8\mathrm{s}$ and then maintains the velocity he has reached. A car $C$ leaves the same road junction as $M$ at time $t=3\mathrm{s}$. The car accelerates from rest to $30\mathrm{m}/\mathrm{s}$ in $20\mathrm{s}$ and then maintains a velocity of $30\mathrm{m}/\mathrm{s}$. $C$ passes $M$ as they both pass a pedestrian.

(i) On the same diagram, sketch velocity-time graphs to illustrate the motion of $M$ and $C$.

(ii) Find the distance of the pedestrian from the road junction.

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Background and Explanation

This problem involves analyzing motion with constant acceleration followed by constant velocity. You will need to apply the SUVAT equations of motion and understand that the area under a velocity-time graph represents displacement. The key challenge is establishing the correct time intervals for each phase of motion, particularly since the car starts 3 seconds later than the motorcycle.

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Solution

Part (i): Velocity-Time Graphs

The velocity-time graphs should show the following characteristics:

For Motorcyclist $M$:

• A straight line starting from the origin $(0,0)$ with a positive gradient, reaching the point $(8,24)$ since $v=3\times 8=24m/s$ at $t=8\mathrm{s}$.
• A horizontal line at $v=24m/s$ for all $t>8\mathrm{s}$.

For Car $C$:

• No motion from $t=0$ to $t=3\mathrm{s}$ (stays on the time axis).
• A straight line from $(3,0)$ to $(23,30)$, since it takes $20\mathrm{s}$ to reach $30m/s$ (from $t=3$ to $t=23$).
• A horizontal line at $v=30m/s$ for all $t>23\mathrm{s}$.

The graph for $C$ starts later but eventually overtakes $M$ because it reaches a higher final velocity ($30m/s$ vs $24m/s$).

Part (ii): Distance of the Pedestrian

To find where the car passes the motorcycle, we calculate the distance traveled by each vehicle as a function of time and set them equal.

Step 1: Analyze the Motion of Motorcycle $M$

Phase 1: Acceleration ($0\le t\le 8$)

Using the equation $v=u+at$: $v_f=0+3\times 8=24m/s$

Using the equation $s=ut+\frac{1}{2}a{t}^2$ for the distance traveled during the first 8 seconds: $S_{1M}=0\times 8+\frac{1}{2}\times 3\times 8^2=\frac{1}{2}\times 3\times 64=96\mathrm{m}$

Phase 2: Constant Velocity ($t>8$)

For times after $t=8\mathrm{s}$, the motorcycle travels at a constant $24m/s$. The additional distance traveled is: $S_{2M}=24(t-8)=24t-192$

Total Distance for $M$: $S_M=96+(24t-192)=24t-96$

Step 2: Analyze the Motion of Car $C$

The car starts at $t=3\mathrm{s}$ and accelerates for $20\mathrm{s}$, so it reaches $30m/s$ at $t=23\mathrm{s}$.

Phase 1: Acceleration ($3\le t\le 23$)

Using the average velocity formula $s=\frac{u+v}{2}\times t$ for the distance traveled during the $20\mathrm{s}$ acceleration period: $S_{1C}=\frac{0+30}{2}\times 20=15\times 20=300\mathrm{m}$

Phase 2: Constant Velocity ($t>23$)

For times after $t=23\mathrm{s}$, the car travels at a constant $30m/s$. The time spent at this velocity is $(t-23)$, so the additional distance is: $S_{2C}=30(t-23)=30(t-20-3)=30t-690$

Total Distance for $C$: $S_C=300+(30t-690)=30t-390$

Step 3: Find the Time When $C$ Passes $M$

When the car passes the motorcycle at the pedestrian's location, both have traveled the same distance from the junction. Therefore, we set $S_C=S_M$:

$30t-390=24t-96$

Solving for $t$: $30t-24t=390-96$ $6t=294$ $t=\frac{294}{6}=49\mathrm{s}$

This occurs at $t=49\mathrm{s}$, which is valid since both vehicles have finished accelerating by this time ($49>23$ and $49>8$).

Step 4: Calculate the Distance of the Pedestrian

Substitute $t=49\mathrm{s}$ into the distance equation for the motorcycle:

$S_M=96+24(49-8)$ $S_M=96+24\times 41$ $S_M=96+984$ $S_M=1080\mathrm{m}$

(Verification using the car's equation: $S_C=30(49)-390=1470-390=1080\mathrm{m}$)

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Key Formulas or Methods Used

• SUVAT equations for constant acceleration:
  • $v=u+at$ (final velocity)
  • $s=ut+\frac{1}{2}a{t}^2$ (displacement with acceleration)
  • $s=\frac{u+v}{2}t$ (displacement using average velocity)
• Constant velocity motion: $s=vt$
• Piecewise distance functions: Combining distances from different motion phases to create a total distance equation as a function of time
• Intersection of distance functions: Setting equal the distances traveled by two objects to find when they meet

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Summary of Steps

1. Calculate $M$'s motion: Find the velocity reached after $8\mathrm{s}$ ($24m/s$) and the distance covered during acceleration ($96\mathrm{m}$), then express total distance as $S_M=24t-96$ for $t>8$.
2. Calculate $C$'s motion: Determine the distance covered during the $20\mathrm{s}$ acceleration ($300\mathrm{m}$) and express total distance as $S_C=30t-390$ for $t>23$.
3. Equate distances: Set $S_C=S_M$ to find the time when the car overtakes the motorcycle: $30t-390=24t-96$.
4. Solve for time: Rearrange to get $6t=294$, yielding $t=49\mathrm{s}$.
5. Find distance: Substitute $t=49$ back into either distance equation to obtain $1080\mathrm{m}$.