Problem 2:
Data:
Car:
Vi=0 Vf = 22.3 m/s ; a = 4.02
m/s2
Bike:
Vi=0 Vf = 8.94 m/s ; a = 5.81
m/s2
So let's calculate how long it takes the bike to reach its cruising
speed.
t = (vf-vi)/a ; t = 8.94/5.81 ; t = 1.539
sec
Now let's calculate the distance it covered during that time.
d = 1/2at2 ; d = 1/2(5.81)(1.539`2) ;d = 6.88
m
Now calculate how long it took the car to reach the bike's
cruising speed.
t = (vf-vi)/a ; t = 22.3/4.02 ; t = 5.55
sec
Calculate the distance traveled during this time
d = 1/2 at2 ; d = 1/2(4.02)(5.552) ;d = 61.91
m
At 1.539 sec and a distance of 6.88 m, the car is not accelerating any more so its distance from the start line can be calculated by the equation d = 8.94 t - 6.88
The distance the car travels from the start line after it reaches the bike's cruising speed is d = 8.94 t + 1/2(4.02)t2 - 61.91 since the car is still accelerating.
Since the total distances will be the same when the car catches up to the bike we substitute the first equation into the second equation.
8.94 t - 6.88 = 8.94 t + 1/2(4.02) t 2 - 61.91
Solve for t.
t = 5.23 s which is the time from the car reaching
the bike's final velocity which took 5.55 s. Add these times
together
5.23 + 5.55 = 10.78 sec
Instructions for PHY 2048 Problem Set (PSET): (1) Please NEATLY write your name and your solutions....
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