Question

A bike race is conducted on a circular track of radius $60\text{ m}$. If the slowest bike, weighing $120\text{ kg}$, moved around the track at a constant speed of $108\text{ kmph}$, then the time taken by it to complete one lap and its acceleration are respectively

• A. $3.49\text{ s}$, $1.11\text{ ms}^{-2}$
• B. $12.56\text{ s}$, $15\text{ ms}^{-2}$
• C. $3.49\text{ s}$, $0\text{ ms}^{-2}$
• D. $2\text{ s}$, $9.8\text{ ms}^{-2}$

Hint:

Always convert speed from kmph to $\text{ms}^{-1}$ early by multiplying by $\frac{5}{18}$ to avoid unit mismatched calculations.

Answer 1

The Correct Option is B

Step 1: Change the speed to SI units.
To go from km/h to m/s we multiply by $\tfrac{5}{18}$. \[ v = 108 \times \frac{5}{18} = 30 \text{ m s}^{-1} \]
Step 2: Pick the lap-time formula.
One full lap is the circumference $2\pi r$. Time for one lap is distance over speed: $T = \dfrac{2\pi r}{v}$.

Step 3: Put in the values.
\[ T = \frac{2\pi (60)}{30} = 4\pi \approx 12.56 \text{ s} \]
Step 4: Think about the acceleration.
The speed stays the same, but the direction keeps changing on a circle. A change in direction still means there is an acceleration.

Step 5: Use the centripetal formula.
This acceleration points to the centre and equals $a = \dfrac{v^{2}}{r}$. \[ a = \frac{30^{2}}{60} = \frac{900}{60} = 15 \text{ m s}^{-2} \]
Step 6: Match the option.
So the lap time is $12.56$ s and the acceleration is $15$ m s$^{-2}$. This is option 2. The acceleration is not zero because the direction keeps turning.
\[ \boxed{T = 12.56 \text{ s},\ a = 15 \text{ m s}^{-2}} \]