A motorbike at a constant rate from a standing start. After 1.2s it is travelling 6.0m/s. how much time will have elapsed starting from rest before the bike is moving with a speed of 15m/s
vf=at a= v/t=6/1.2=5m/s^2
t=vf/a=15/5 sec
V = Vo + a*t = 6.
0 + a*1.2 = 6,
a = 5 m/s^2.
V = Vo + a*t = 15.
0 + 5*t = 15,
t =
To find out how much time will have elapsed starting from rest before the bike is moving with a speed of 15 m/s, we can use the equation of motion that relates velocity, time, and acceleration: v = u + at.
In this equation:
- v is the final velocity (15 m/s)
- u is the initial velocity (0 m/s, as the bike starts from rest)
- a is the acceleration (which we assume to be constant)
- t is the time taken to reach the final velocity
Since the acceleration is constant, we can use the equation v = u + at to find the time needed for the bike to reach a speed of 15 m/s. Rearranging the equation, we get:
t = (v - u) / a
Since u = 0, the equation simplifies to:
t = v / a
Now, we need to know the acceleration. Unfortunately, we do not have the acceleration value given in the question. However, we can determine it using the information provided.
We are given the speed of the bike after 1.2 seconds, which is 6.0 m/s. By substituting these values into the equation of motion, we can solve for the acceleration:
v = u + at
6.0 = 0 + a * 1.2
Simplifying the equation:
6.0 = 1.2a
To solve for a, we divide both sides of the equation by 1.2:
a = 6.0 / 1.2
a = 5.0 m/s^2
Now that we have the acceleration, we can substitute it back into the equation t = v / a:
t = 15 / 5.0
t = 3.0 seconds
Therefore, starting from rest, it will take 3.0 seconds for the motorbike to reach a speed of 15 m/s.