Answer:
A.) 8 seconds
B.) 16 seconds
C.) 48 m
Explanation:
Given that a cyclist traveling at constant speed of 12 m/s
and the bus accelerates uniformly at 1.5ms²
A.) The bus has the following parameters
Acceleration a = 1.5 m/s^2
Initial velocity U = 0. Since the bus is starting from rest.
Final velocity V = 12 m/s
Use equation one of linear motion.
V = U + at
Substitute V, U and a into the formula
12 = 0 + 1.5t
1.5t = 12
t = 12/1.5
t = 8 seconds
Therefore, the bus reach the same speed as the cyclist at 8 seconds.
B.) For the cyclist moving at constant speed, acceleration a = 0. Using second equation of motion
h = Ut + 1/2at^2
Since a = 0, the equation is reduced to:
h = Ut.
Also, for the bus,
h = Ut + 1/2at^2
Equate the two equations since the h is the same
Ut = Ut + 1/2at^2
Substitute all the parameters into the formula
12t = 0 + 1/2 × 1.5t^2
12t = 0.75t^2
0.75t = 12
t = 12/0.75
t = 16 seconds
Therefore, the bus takes 16 seconds to catch the cyclist
C.) Use third equation of linear motion.
V^2 = U^2 + 2as
Where s = distance
Substitute V, U and a into the formula
12^2 = 0 + 2 × 1.5 S
144 = 3S
S = 144/3
S = 48 m
