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)=?t+?t2, with ?=2.60m/s and ?=1.20m/s2. The distance of B from the starting point is xB(t)=?t2??t3, with ?=2.80m/s2 and ?=0.20m/s3.

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.

Answer 1

To find out which car is ahead, just find out the values of xa and xb at t = 0, i.e.

xa(0) = a(0) + b(0)^2
= 0 m

xb(0) = c(0)^2 - d(0)^3
= 0 m

So, we see that both of the cars are at x=0 at time t=0.


Part B: We know that the cars are at the same point at t=0, but to find any other time at which both cars are at the same point, we will use the following relation,

at + bt^2 = ct^2 - dt^3
dt^3 + (b - c)t^2 + at = 0
taking t common,
dt^2 + (b-c)t + a = 0
(0.2)t^2 + (1.2 - 2.8) t + 2.6 = 0
(0.2)t^2 + (-1.6)t + 2.6 = 0
t^2 -8t + 13 = 0
solve this quadratic equation and you will get the required value of t.


Part C: I suggest, it means that their distance from each other remains constant. i.e. their velocities are equal~

xa(t) = at + bt^2
va(t) = dxa/dt = 2bt + a

xb(t) = ct^2 - dt^3
vb(t) = 2ct - 3dt^2

Now, we know from the given condition that,
va = vb

2bt + a = 2ct -3dt^2
3dt^2 + 2(b-c)t +a = 0

Now, simplify this quadratic equation to get the value of t. and that will be your answer(s).


Part D: Acceleration should be same~

Aa(t) = dva/dt = 2b

Ab(t) = dvb/dt = 2c -6dt

Aa = Ab
2b = 2c - 6dt
b = c - 3dt
3dt = c - b
t = (c - b)/3d