Question

A particle moves along a straight line with a velocity in meters per second given by v = 12 - 3t + 8t, where t is in seconds. When t = 0s the position of the particle is +5m. Determine: A. The equation for position as a function of time.B. The equation for the acceleration as a function of time.C. The acceleration of the particle at t = 6 s.D. The net displacement, ∆s, from t = 0 to t = 6 s.E. The total distance traveled, sT, from t = 0 to t = 6 s.F. The average velocity, vavg, from t = 0 to t = 6 s.

Answer 1

Answer:

Explanation:

Given the velocity of a particle modeled by the equation

v = 13-3t+8t² where t is in seconds

Given t = 0 and position s = +5m

A) To get the position as a function of time, we will integrate the function with respect to t ad shown;

v = 13-3t+8t²

S = ∫13-3t+8t² dt

S = 13t-13t²/2+8t³/3 + C

at t = 0 and S = +5m

5 = 13(0)-13(0)²/2+8(0)³/3+C

5 = 0-0+0+C

C = 5

Substituting c = 5 into the displacement function

S = 13t-13t²/2+8t³/3 + C

S = 13t-13t²/2+8t³/3 + 5

B) acceleration is the change in velocity with respect to time.

a = dv/dt

Given v = 13-3t+8t²

a= dv/dt = -3+16t

a = 16-3t

C) acceleration at t = 6s is derived by plugging in t = 6 into the resulting equations in (B)

a = 16-3t

a = 16-3(6)

a = 16-18

a = -2m/s²

D) net displacement from t = 0 to t = 6s

At t = 0:

S(0) = 13(0)-13(0)²/2+8(0)³/3 + 5

S(0) = 0+5

S(0) = 5m

At t = 6s

S(6) = 13(6)-13(6)²/2+8(6)³/3 + 5

S(6) = 78-234+576+5

S(6) = 425m

Net displacement from t = 0s to t = 6s is s(6)-s(0)

= 425-5

= 420m

E) Total distance travelled D = S(6)+S(0)

= 425+5

= 430m

F) Average velocity = ∆S/∆t

Average velocity = S(6)-S(0)/6-0

Average velocity = 425-5/6

Average velocity = 420/6

Average velocity = 70m/s