The figure below shows the acceleration as a function of time for an object.
(a) If the object starts from rest at t = 0, what is the velocity of the object as a function of time?
for 0s < t < 10s v(t) = m/s
for 10s < t < 26s v(t) = m/s
(b) If the object instead has a velocity of −40 m/s at t = 0, how does your result for part (a) change?
for 0 < t < 10s v(t) = m/s
for 10s < t < 26s v(t) = m/s
figure?
To find the velocity of the object as a function of time, we need to integrate the acceleration with respect to time.
(a) Assuming the object starts from rest at t = 0, the velocity of the object can be found by integrating the acceleration function between 0s and the given time intervals.
For 0s < t < 10s:
To find the velocity function for this time interval, we need to integrate the given acceleration function. Integrate the acceleration function with respect to time to get the velocity function.
Let's denote the integral of acceleration with respect to time as V(t) (velocity function).
∫ a(t) dt = V(t) (1)
Integrating acceleration function within this time interval:
∫ 0 dt = ∫ a(t) dt
0 = ∫ a(t) dt
Therefore, the velocity function for 0s < t < 10s is v(t) = 0 m/s.
For 10s < t < 26s:
Similarly, integrate the given acceleration function within this time interval to find the velocity function:
∫ a(t) dt = V(t) (2)
∫ a(t) dt = ∫ C dt, where C is a constant of integration
Integrating acceleration function within this time interval:
∫ a(t) dt = ∫ C dt
Therefore, the velocity function for 10s < t < 26s is v(t) = C t + D m/s, where C and D are constants of integration.
(b) If the object instead has a velocity of -40 m/s at t = 0, the answer for part (a) changes as follows:
For 0 < t < 10s:
Now, the initial velocity is -40 m/s instead of 0 m/s. Using equation (1), we need to find the velocity function by integrating the acceleration function between 0s and 10s.
∫ a(t) dt = V(t)
Similarly, integrating the acceleration function within this time interval:
∫ a(t) dt = V(t) = ∫ C dt
Therefore, the velocity function for 0 < t < 10s will be v(t) = C t + D m/s, where C and D are constants of integration.
For 10s < t < 26s:
The same equation (2) will apply for this time interval. Integrate the given acceleration function within this time interval to find the velocity function:
∫ a(t) dt = V(t) = ∫ C dt
Therefore, the velocity function for 10s < t < 26s will still be v(t) = C t + D m/s, where C and D are constants of integration.
To fully determine the velocity functions for each time interval, the constants of integration (C and D) need to be determined. This information is not provided in the question, so additional information is needed to find the actual velocity functions.