Answer:
Part a)
\(t = 2.96 s\)
Part b)
\(d = 1.55 \times 10^{-3} miles\)
Explanation:
Part a)
cruising velocity of bicycle = 19 mi/h
acceleration = 14.50 mi/h/s
For the motion of car
cruising velocity of car = 58 mi/h
acceleration = 10 mi/h/s
Both start from rest
Now we have
time taken by bicycle to reach its cruising speed is given as
\(t = \frac{\Delta v}{a}\)
\(t = \frac{19 - 0}{14.50} = 1.31 s\)
Time taken by car to reach its cruising speed
\(t_2 = \frac{58- 0}{10} = 5.8 s\)
now the distance moved by the car = distance moved by bicycle
till the time cycle moving in front of car
distance moved by the car = \(\frac{1}{2}at^2\)
distance moved by bicycle = \(\frac{1}{2}at_1^2 + v(t - t_1)\)
\(\frac{1}{2}(10)(t^2) = \frac{1}{2}(14.50)(1.31^2) + 19(t - 1.31)\)
\(5t^2 = 12.44 + 19t - 24.89\)
\(t = 2.96 s\)
Part b)
Bicycle will lead the car till the speed of car and speed of bicycle is same
so we have
\(V_{cycle} = V_{car}\)
\(19.0 mi/h = 0 + 10(t)\)
\(t = 1.9 s\)
so the distance between them is given as
\(d = (\frac{1}{2}(14.50)(1.31^2) + 19(1.9 - 1.31)) - (\frac{1}{2}(10)(1.9^2))\)
\(d = (6.56 \times 10^{-3}) - (5.01 \times 10^{-3})\)
\(d = 1.55 \times 10^{-3} miles\)
