Respuesta :
Answer: (a) The bicycle is ahead of the car for 4 s.
(b) The bicycle leads the car by the maximum distance of 55 m.
Explanation:
(a)
Use the equation of the motion to calculate the time taken by the car.
[tex]v=u+at[/tex]
As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec2 .
Put u=0, v=22.3 m/sec and a=4.02 m/sec^2.
[tex]22.3=0+4.02t[/tex]
[tex]t=\frac{22.3}{4.02}[/tex]
t= 5.5 s
Use the equation of the motion to calculate the time taken by the bicycle.
[tex]v=u+at_{1}[/tex]
As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.
Put u=0, v=8.94 m/sec and a=5.81 m/sec^2.
[tex]8.94=0+5.81t_{1}[/tex]
[tex]t_{1}=\frac{8.94}{5.81}
[tex]t_{1}=1.5 s[/tex]
Calculate the time interval for which the bicycle is ahead of the car.
[tex]t-t_{1}= 5.5 s - 1.5s[/tex]
[tex]t-t_{1}= 4s[/tex]
Therefore, the bicycle is ahead of the car for 4 s.
(b)
Use the equation motion to calculate the distance covered by the car.
[tex]S=ut+\frac{1}{2}at^{2}[/tex]
As it is given in the problem, a car speeds up from rest to its cruising speed 22.3 m/sec with constant acceleration 4.02 m/sec^2 .
Put t= 5.5 s, u=0 s and a=4.02 m/sec^2.
[tex]S=(0)t+\frac{1}{2}(4.02)(5.5)^{2}[/tex]
[tex]S= 60.8 m[/tex]
Use the equation motion to calculate the distance covered by the bicycle.
[tex]S_{1}=ut+\frac{1}{2}at^{2}[/tex]
As it is given in the problem, a cyclist speeds up from rest to its cruising speed 8.94 m/sec with constant acceleration 5.81 m/sec^2.
Put t= 1.5 s, u=0 s and a=5.81 m/sec^2.
[tex]S_{1}=(0)t+\frac{1}{2}(5.18)(1.5)^{2}[/tex]
[tex]S_{1}= 5.8 m[/tex]
Calculate the maximum distance covered by the bicycle to lead the car.
[tex]S-S_{1}=60.8-5.8=55m[/tex]
Therefore, the bicycle leads the car by the maximum distance of 55 m.
