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
