Final answer:
The maximum velocity of the motorcycle occurs at t = 6.25 seconds and is equal to 7.8125 m/s, is the acceleration of a motorcycle moving on a straight line is given by a = 1.0* t – 0.16 t2, where t in s and a in m/s2. The motorcycle is at rest at the origin at time t = 0.
Step-by-step explanation:
To find the maximum velocity of the motorcycle, we need to determine when the acceleration becomes zero.
Given that the acceleration is given by the equation a = 1.0t - 0.16t2, we can set the acceleration equal to zero:
0 = 1.0t - 0.16t2
0 = t(1.0 - 0.16t)
Using the zero-product property, we can set each factor equal to zero:
t = 0 or 1.0 - 0.16t = 0
The first solution t = 0 represents the initial time when the motorcycle is at rest.
Plugging the second solution into the equation, we can solve for t:
1.0 - 0.16t = 0
0.16t = 1.0
t = 6.25
Therefore, the maximum velocity of the motorcycle occurs at t = 6.25 seconds.
Plugging this value of t into the equation for the velocity, we can find the maximum velocity:
v = ∫(1.0t - 0.16t2) dt
v = 0.5t2 - (1/3)(0.16t3) + C
Since the motorcycle is at rest at t = 0, C = 0.
Therefore, the velocity function becomes:
v = 0.5t2 - (1/3)(0.16t3)
Substituting t = 6.25:
v = 0.5(6.25)2 - (1/3)(0.16)(6.25)3
v = 15.625 - 7.8125
v = 7.8125 m/s