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Calculus Help?
The acceleration of a particle in one dimensional motion at time t is given by a(t)=6t-1. Given that the initial velocity of the particle is 0, find the average velocity of this particle over the first 4 seconds.
Pls explain thank you!!
2 Answers
- 3 years ago
a(t)=6t-1
a(t)= derivative of v(t)
v(t)=3t^2-t
v(t)=derivative of s(t)= distance vs. time function
s(t)=t^3-t^2/2
Final distance= 4^3-(4)^2/2=56
Initial Distance=0
Average Velocity=(56-0)/4=14
- ?Lv 73 years ago
If a(t) = 6t - 1, then by integration and using the given v(0)=0
v(t) = 3 t^2 - t
One more integration gives distance
s(t) - s(0) = t^3- t^2 / 2
The average velocity then is
( s(4)-s(0) ) / 4 = 56/4 = 14.
