1 answer

The position of a particle along a straight-line path is defined by s=(t3−6t2−15t+7) ft, wheret is...

Question:

The position of a particle along a straight-line path is defined by s=(t3−6t2−15t+7) ft, wheret is in seconds.

Part A: Determine the total distance traveled when t = 8.3 s .

Part B: What are the particle's average velocity at the time given in part A?

Part C: What are the particle's average speed at the time given in part A?

Part D: What are the particle's instantaneous velocity at the time given in part A?

Part E: What are the particle's acceleration at the time given in part A?


Answers

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

Part B:

Average velocity = Displacement / time = 33.95 / 8.3 = 4.1 ft /s

Part C:

Aaverage speed = Distance covered / time = 233.95 / 8.3 = 28.19 ft/s

Part D:

Instantaneous velocity = dS/dt at t=8.3

=3t2 - 12t -15

= 92.07 ft/ s

Part E:

Acceleration = dV/dt at t= 8.3

= 6t-12 at t=8.3

= 37.8 ft/s2

  

.

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