A motorcycle has a constant acceleration of 4.06 m/s2. Both the velocity and acceleration of the motorcycle point in the same direction. How much time is required for the motorcycle to change its speed from (a) 39.4 to 49.4 m/s, and (b) 69.4 to 79.4 m/s?
a. V = Vo + at.
t = (V-Vo)/a = (49.4-39.4)/4.06 =
2.463
To solve this problem, we can use the kinematic equation:
v = u + at,
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration,
t is the time.
a) Let's first solve for the time required to change the speed from 39.4 m/s to 49.4 m/s. Here,
u = 39.4 m/s,
v = 49.4 m/s,
a = 4.06 m/s^2.
We can rearrange the kinematic equation to solve for time:
t = (v - u) / a.
Substituting the values:
t = (49.4 - 39.4) / 4.06 = 2.4657 seconds (rounded to four decimal places).
Therefore, it takes approximately 2.4657 seconds for the motorcycle to change its speed from 39.4 m/s to 49.4 m/s.
b) Now, let's solve for the time required to change the speed from 69.4 m/s to 79.4 m/s. Using the same equation:
u = 69.4 m/s,
v = 79.4 m/s,
a = 4.06 m/s^2.
Again, substituting the values into the equation:
t = (79.4 - 69.4) / 4.06 = 2.4630 seconds (rounded to four decimal places).
Therefore, it takes approximately 2.4630 seconds for the motorcycle to change its speed from 69.4 m/s to 79.4 m/s.