The position of a particle along a straight-line path is defined by S =(t3−6t2−15t+7) ft, wheret is in seconds
when t =0 then S= 7 ft
If we differentiate S with respect to time we get velocity of the particle.
which is V = 3t2 - 12t -15 ft/s
at t=0 V= -15 ft/s
and V= 0 at t= 5 second i.e the particle will reverse the direction of its initial velocity at this time.
Part A: the total distance traveled when t = 8.3 s will be the sum of the magnitude of displacement of particle from 0 to 5 seconds and from 5 to 8.3 seconds.
S at t= 0 is 7 ft.
S at t= 5 is -93
S at t= 8.3 is 40.95 ft
Displacement from t=0 to t=5 is equal to 100 ft
Displacement from t=5to t=8.3 is equal to 133.95 ft
Hence distance traveled = 133.947 ft + 100 ft = 233.95 ft
Average velocity = Displacement / time = 33.95 / 8.3 = 4.1 ft /s
Aaverage speed = Distance covered / time = 233.95 / 8.3 = 28.19 ft/s
Instantaneous velocity = dS/dt at t=8.3
=3t2 - 12t -15
= 92.07 ft/ s
Acceleration = dV/dt at t= 8.3
= 6t-12 at t=8.3
= 37.8 ft/s2