suppose the velocity of a moving particle is v(t)=t^2-11t+24 ft/sec find.
a) net distance traveled in the first 10 seconds
b) total distance traveled in 10 seconds

Question

suppose the velocity of a moving particle is v(t)=t^2-11t+24 ft/sec find.
a) net distance traveled in the first 10 seconds
b) total distance traveled in 10 seconds

mani_jha · Answer

For calculating total distance
dx/dt=t^2-11t+24
dx=(t^2-11t+24)dt(Equation 1)
Integrate from O to 10.
For calculatng net distance, observe that
$$t ^{2}-11t+24=(t-3)(t-8)$$
Speed becomes 0 at t=3 and t=8.
after t=3, the function becomes negative(put t=4 or 5)
The body moves in opposite direction, until speed becomes zero again at t=8.
To find the net distance calculate integral of equation 1 from 0 to 3 first. Then calculate integral from 3 to 8. Subtract the two.
Now, after t=8, function becomes positive. Find the integral from 8 to 10, and add this to the previous result.
I think its correct. Pls tell me if you have problem understanding

rmrjr22 · Answer

Thanks! I believe i Got it.