Two cars, A and B, travel in a straight line. The distance of A from the startin

Question

Two cars, A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by XA (t)=at+Bt^2, with a=2.60 m/s and B=1.20 m/s^2. The distance of B from the starting point is XB (t)= lt^2-kt^3 , with l=2.80 m/s^2 and k=0.20 m/s^3 .


Part A
Which car is ahead just after they leave the starting point?

Car A
Car B


Part B
At what time(s) are the cars at the same point?
Express your answer numerically. If there is more than one answer, enter each answer separated by a commas.


Part C
At what time(s) is the distance from A to B neither increasing nor decreasing?
Express your answer numerically. If there is more than one answer, enter each answer separated by a comma.


Part D
At what time(s) do A and B have the same acceleration?
Express your answer numerically. If there is more than one answer, enter each answer separated by a comma.

Explanation / Answer

(1)given:-
=>XA(T)=aT+bT^2
=>XA(T) = 2.60T + 1.20T^2 ----------(i)
=>At T = 0
=>XA(0) = 0
& given:-
=>XB(t)=qt^2 -wt^3
=>XB(t) = 2.80t^2 - 0.20t^3
=>at t = 0
=>XB(0) = 0
=>initially both are at the same starting point.
(2) Let at t = m, XA(T) = XB(t)
=>2.60m + 1.20m^2 = 2.80m^2 - 0.20m^3
=>0.20m^3 - 1.60m^2 + 2.60m = 0
=>m[0.20m^2 - 1.60m + 2.60] = 0
=>either m = 0 or [0.20m^2 - 1.60m + 2.60] = 0
=>either m = 0 or m = [-(-1.60) +/- ?{(-1.60)^2 - 4 x 0.20 x 2.60}]/[2x0.20]
=>either m = 0 or m = [1.60 +/- 0.69]/[0.40]
=>either m = 0 or m = 5.73 or 0.77 sec
(3) By V = ds/dT
=>V = a + 2bT
=>By A = dV/dT
=>A = 2b = 2 x 1.20 = 4.20 m/s^2
similarly a = 2q - 6wt
=>a = 2 x 2.80 - 6 x 0.20t
=>a = 5.60 - 1.20t m/s^2
=>But given a = A
=>4.20 = 5.60 - 1.20t
=>t = 1.17 sec

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