Two go karts are racing,Kart XJ moves with a speed of 20mph,when kart ZX is moving at 30mph and about to?

Question

overtake kart XJ,however kart XJ then accelerates at a constant of 0.25mph each second until it passes ZX. How long will ZX be ahead XJ?

Been having trouble solving this one,please show in as detailed as possible. Thankx for the help everyone.

Answer 1

So the initial relative position is zero.
The initial relative velocity (v0) is 10 mph
The constant acceleration (a) is 0.25 mph / second

This is just like throwing a ball straight up and down from the ground. The numbers and units are just a bit different.

The time for the velocities to equalize is the velocity difference over the acceleration: v0/a
Then it takes an equal time to get back to even.
So your answer is 2 v0 / a. Plugnchug.

Answer 2

the question is a bit confusing, but I think I got it
given that at t=0 both have the same position, with ZX moving at 30 mph and XJ moving at 20 mph. ZX passes XJ the instant that XJ begins to accelerate. The solution is when they are at the same position again.
XJ(t)=v1*t+.5*a*t^2
were v1=20 mph
and a=0.25 mph/s
and ZX(t)=v2*t
where v2=30 mph

solve for t when XJ(t)=ZX(t)
v1*t+.5*a*t^2=v2*t
one solution is t-0, which is a boundary condition as stated in the question.
the other is
t=2*(v2-v1)/a
or
t=2*(10)/.25 seconds
t=80 seconds

j

Answer 3

Let t be the time and S be the distance at which J passes Z,
S = average velocity x time = [v +u] *t /2
Distance traveled by Z during this time is 30t.
Since the distances are equal, [v +u] t /2 = 30 t.
v = 60 - 20 = 40 mph.
To find the time t = [v - u]/a = 20 [mph] / 0.25 [mph /s]
t = 80 s. = 1.33 hour.
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Distance traveled by Z = 30 x 1.33 = 39 miles
Distance traveled by J = [40 + 20] *1.33 / 2= 39 miles.
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