Question

Two cyclist race against the clock the clock in a 50.0km cross-country route. Cyclist A travels at a constant speed of 40.0km/h. Cyclist B started 15.0 minutes after cyclist A but manages to catch up at the finish line. What is the speed of cyclist B assuming that his speed is constant?

Answer 1

Answer:

Speed of B is 50km/h

Explanation:

for A:

v = 40km/h

s = 50km/h

so, t = s÷v

t = 50÷40 = 1.25

t = 1h 15min

for B:

t = 1h

s = 50km

v = s÷t

v = 50÷1

v = 50km/h

Answer 2

Given: Two cyclists race against the clock in a 50.0 km cross-country route. Cyclist A travels at a constant speed of 40.0 km/h. Cyclist B started 15.0 minutes after cyclist A but manages to catch up at the finish line.

To find: The speed of cyclist B.

Solution:

• If cyclist A travels 50 km at a constant speed of 40 km/h, the time taken by him can be calculated by the formula,

       $s = \frac{d}{t}$

• Here, s is the speed of the cyclist, d is the distance travelled and t is the time taken.
• On replacing the terms with the values given in the question,

        $40 km/h = \frac{50 km}{t}$

        $t = 1.25 h$

• Cyclist B starts 15 minutes after cyclist A, that is, 0.25 hours after cyclist A. So, the time taken by him is calculated as,

        $t = 1.25 - 0.25$

        $t = 1 h$

• Now, the speed of cyclist B is calculated as,

        $s = \frac{50 km}{1 h}$

           $= 50 km/h$

Therefore, the speed of cyclist B is 50 km/h.