If anyone could help me with these problems it would be greatly appreciated, I've tried to do them but I don't get how I can set up the equation with enough variables to make it work.
1. A bike rider pedals with constant acceleration to reach a velocity of 7.5 m/s over a time of 4.2 s. During the period of acceleration, the bike's displacement is 19 m. What was the initial velocity of the bike?
2. Engineers are developing new types of guns that might someday be used to launch satellites as if they were bullets. One such gun can give a small object a velocity of 2.9 km/s while moving it through a distance of only 2.1 cm.
(a) What acceleration does the gun give this object?
b) Over what time interval does the acceleration take place?
1. A bike rider pedals with constant acceleration to reach a velocity of 7.5 m/s over a time of 4.2 s. During the period of acceleration, the bike's displacement is 19 m. What was the initial velocity of the bike?
2. Engineers are developing new types of guns that might someday be used to launch satellites as if they were bullets. One such gun can give a small object a velocity of 2.9 km/s while moving it through a distance of only 2.1 cm.
(a) What acceleration does the gun give this object?
b) Over what time interval does the acceleration take place?
-
Use equations of kinematics for this.
s = vt + 0.5at^2
s = 0.5(V + v)t
V = v + at
V^2 = v^2 + 2as
Where:
s = displacement
V = final velocity
v = initial velocity
a = acceleration
t = time
For 1. I will use S=0.5(V + v)t to find v.
S = 19m
V = 7.5m/s
t = 4.2s
19 = 0.5(7.5 + v)4.2
v = (19/2.1) - 7.5 = 1.55m/s
For 2a, I would use V^2 = v^2 + 2as
V = 2.9km/s = 2900m/s
v = 0km/s = 0m/s
s = 2.1cm = 0.021m
Note: because there is a difference between units in the information provided, we must convert to a common unit)
(2900)^2 = (0)^2 + 2(0.021)a
a = 200238095m/s^2 = 200238.095km/s^2
For 2b, I would use s = 0.5(V + v)t
s = 2.1cm = 0.021m
V = 2.9km/s = 2900m/s
v = 0m/s
0.021 = 0.5 (2900 + 0)t
t = 0.021/(0.5 x 2900) = 3.62x10^-6s
Do please check over this and make sure the numbers are correct. I cannot guarantee that my calculations are perfect!
s = vt + 0.5at^2
s = 0.5(V + v)t
V = v + at
V^2 = v^2 + 2as
Where:
s = displacement
V = final velocity
v = initial velocity
a = acceleration
t = time
For 1. I will use S=0.5(V + v)t to find v.
S = 19m
V = 7.5m/s
t = 4.2s
19 = 0.5(7.5 + v)4.2
v = (19/2.1) - 7.5 = 1.55m/s
For 2a, I would use V^2 = v^2 + 2as
V = 2.9km/s = 2900m/s
v = 0km/s = 0m/s
s = 2.1cm = 0.021m
Note: because there is a difference between units in the information provided, we must convert to a common unit)
(2900)^2 = (0)^2 + 2(0.021)a
a = 200238095m/s^2 = 200238.095km/s^2
For 2b, I would use s = 0.5(V + v)t
s = 2.1cm = 0.021m
V = 2.9km/s = 2900m/s
v = 0m/s
0.021 = 0.5 (2900 + 0)t
t = 0.021/(0.5 x 2900) = 3.62x10^-6s
Do please check over this and make sure the numbers are correct. I cannot guarantee that my calculations are perfect!