Question

Two cars are starting. Both are accelerating with a constant acceleration. After t=15s car 2 is 200m ahead of car 1. How large is the velocity of both cars?

Answer 1

u want individual velocities???

Answer 2

Yes. Of car 1 and car 2.

Answer 3

Well you have this equation\[x _{f}=x _{i}+v _{i}t+\frac{ 1 }{ 2 }at ^{2}\]So you have two equations:\[x _{f _{1}}=V _{i _{1}}15s+\frac{ 1 }{ 2 }a _{1}(15s)^{2}\]\[200m+x _{f _{1}}=v _{i _{2}}15s+\frac{ 1 }{ 2 }a _{2}(15s)^{2}\]So if you melt the to together through xf1 \[v _{i _{1}}15s+\frac{ 1 }{ 2 }a _{1}(15s)^{2}=v _{i _{2}}15s+\frac{ 1 }{ 2 }a _{2}(15s)^{2}-200m\]That's as far as I can go unless tha problem says that both cars started with the same velocity or had the same acceleration. Does it?

Answer 4

Now,both cars are starting so their initial velocities are zero.
now\[s=ut+1/2at ^{2}\] u = 0 let s = x (distance)
so, a =2x/225

Case 2
The second car covers 200m more than the first car
So,after calculation, its acceleration would be 2x/225+400/225

This clearly implies that the acceleration of second car was more than the first car by approx.\[1.7m/s ^{2}\] (400/225)
With this acceleratiomn, it covers a distance of 200m
Now\[v ^{2}-u ^{2}=2as\]
\[v ^{2}=2*1.7*200=\sqrt[2]{680}=26.07m/s\]
Similarly you can calculate the velocity of the first car by getting the distance covered by the second car,then subtracting 200 to get the distance of the first car and then finding the velocity .

Answer 5

Not enough data to calculate individual velocities\[S_1=a_1t^2/2\]\[S_2=a_2t^2/2\]
\[S_2-S_1=200=(a_2-a_1)t^2/2\rightarrow a_2-a_1=400/t^2\]\[V_1=a_1t\]\[V_2=a_2t\]
\[V_2-V_1=(a_2-a_1)t \rightarrow V_2-V_1=(400/t^2)·t=400/t=400/15\]That means that after 15 seconds, the relation between both velocities is, in meters per second:
\[V_2=V_1+400/15\]

Answer 6

And at any moment, the relation between both velocities is given by:
\[V_2=V_1+(400/225)t\]