Cars A and B are racing each other along a straight path in following manner: Car A has head start and is a distance dA beyond starting line at t=0. The starting line is at x=o. Car A travels at constant speed vA. Car B starts at starting line but has better engine than car A and travels at constant speed vB (which is greater than vA). How long after Car B started the race will Car B catch up with Car A? How far from Car B's starting line will the cars be when Car B passes Car A?

Da/(Vb-Va)

Is the right answer.

that helped me get a full mark thanks

A test car travels in a straight line along the x axis. The graph in the figure shows the car's position x as a function of time. Find a) the average velocity between t₁ = 0.0 s and t₂ = 6.0 s. b) the instantaneous velocity at points A, D and G. c) The instantaneous speed at points A, D and G. d) the average speed between t₁ = : 0.0 s and t₂ = 6.0 s

To find the time it takes for Car B to catch up with Car A, we'll first calculate the time it takes for Car B to cover the initial distance dA that Car A had as a head start.

Given:
Distance of Car A's head start: dA
Speed of Car A: vA
Speed of Car B: vB (vB > vA)

Let's assume that Car B starts at t=0. At that time, Car A is already ahead by a distance dA.

Distance traveled by Car A after time t is given by:
dA = vA * t

Now, we can determine the time when Car B catches up with Car A. When Car B catches up, the distance both cars have traveled will be equal.

Distance traveled by Car B after time t is given by:
dB = vB * t

Since Car B started at the same point as Car A, its distance traveled will be the initial head start distance dA plus the distance it covers in time t.

dB = dA + vB * t

Setting the two equations equal to each other, we have:

vA * t = dA + vB * t

Rearranging the equation, we get:

vA * t - vB * t = dA

(t * (vA - vB)) = dA

Now, we divide both sides of the equation by (vA - vB):

t = dA / (vA - vB)

This gives us the time it takes for Car B to catch up with Car A.

To find the distance from Car B's starting line when Car B passes Car A, we substitute the value of t we just found into either of the equations for distance:

dB = dA + vB * t

Now, substitute the value of t:

dB = dA + vB * (dA / (vA - vB))

Simplifying the equation, we get:

dB = (dA * (vA/vB)) + dA

Therefore, the cars will be at a distance of dA * (1 + (vA/vB)) from Car B's starting line when Car B passes Car A.

To summarize:

At t=0
car A: x=dA, speed=vA
car B: x=0, speed=vB
When will car B overtake car A and where.

Distance to catch-up = dA
difference in speed = (vB-vA)
Time to catch up, T = dA/(vB-vA)
location where the two cars are side-by-side
= T*vB
= dA*vB/(vB-vA)