A bike rider accelerates constantly to a velocity of 8.0 m/s during 4.2 s. The bike's displacement during the acceleration was 19 m. What was the initial velocity of the bike?
x= 1/2(Vo + V)t
Solve for Vo
Post your work if you get stuck.
can u give the answer
x = 19 m
Vo = unknown
V = 8.0 m/s
t = 4.2 s
To find the initial velocity of the bike, we can use the formula for displacement during constant acceleration:
s = ut + (1/2)at^2
Where:
s = displacement
u = initial velocity
t = time
a = acceleration
Given:
s = 19 m
t = 4.2 s
v = 8.0 m/s (final velocity)
We need to find u.
Since the problem states that the bike accelerates constantly, we can assume the acceleration (a) remains constant during the 4.2 seconds.
Rearranging the formula, we have:
s = ut + (1/2)at^2
19 = u(4.2) + (1/2)a(4.2)^2
Now, since the equation states that the velocity of the bike reaches 8.0 m/s at the end of the 4.2 seconds, we can use the following equation to relate the final velocity (v) to the initial velocity (u) and acceleration (a):
v = u + at
8.0 = u + (a)(4.2)
We can use these two equations to solve for u and a.
Step 1: Solve the second equation for a:
a = (8.0 - u) / 4.2
Step 2: Substitute this value for a in the first equation:
19 = u(4.2) + (1/2)((8.0 - u) / 4.2)(4.2)^2
Now, we can solve this equation for u.
19 = 4.2u + (1/2)(8.0 - u)(4.2)
Simplify the equation:
19 = 4.2u + (1/2)(33.6 - 4.2u)
Expand and combine like terms:
19 = 4.2u + 16.8 - 2.1u
Move the terms involving u to one side and the constants to the other side:
4.2u - 2.1u = 19 - 16.8
Combine like terms:
2.1u = 2.2
Divide both sides by 2.1:
u = 2.2 / 2.1
u = 1.05 m/s
Therefore, the initial velocity of the bike was 1.05 m/s.