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Kinematics one dimension. help

kinematics one dimension. help!!

hi i've had a lot of trouble with this problem and i hope someone can help me.

Car B starts from rest and accelerates at 2.00m/s^2 along a straight road. Car A starts from rest at the same instant 20.0 m behind car B and accelerates 3.00 m/s^2. How far must car A move in order to catch car B? What will be the speeds of car A and car B?

im pretty sure that the time is equal in both cases so t_a=t_b. but it is also asking for distance, and i don't see any equation that requires only acceleration and initial velocity. i have no idea where that 20 m behind goes to.

Can someone explain this problem to me and show me how to do it? I am very confused. :confused: Thanks!!
 
Last edited:
ok so far i got Da(distance of A)=20+Db and Db=1/2at^2. do u just plug in the values and set them equal?
 
nevermind i finally figured the problem out. but i have another one im having trouble with.

Car B is moving at a constant speed of 25m/s along a straight road. Car A starts from rest and moves in the same direction when B is 20.0 m ahead. If A accelerates at a rate of 5 m/s^2, how far must it travel in order to catch B? How long will it take? How fast will A be travelling when it catches B?

Does constant speed mean initial velocity?
can someone explain to me how to do this?

thanks
 
575
2
Start by developing an equation for the speed of A relative to B (pretend B is stationary). Can you take it from there?
 
so i should take that car B has an acceleration of 0? im still confused on what constant speed is - does that mean average velocity, final velocity, or initial velocity?
 
ok i counted the 25m/s as initial velocity and the acceleration 0 for car B. I set car A and B equal using the equation d=vi(t)+1/2(a)(t)^2.

so, 1/2(5)t^2=20+25t
then , 2.5(t)^2-25(t)-20=0 and use the quadratic formula? is that it?
 
395
1
I don't like using formulae. Think of a velocity-time graph. The line representing the movement of Car B will have a gradient of 2. The line representing the movement of Car A will have a gradient of 3. The distance moved by each car is the distance underneath each line.

When Car A catches up with Car B, it will have travelled the same amount of time as Car B, but it will have traversed an extra distance of 20m.

In the graph, because the area underneath each of the lines is a triangle, we can calculate

1/2 x t x 2t + 20 = (1/2 x t x 3t)
t^2 + 20 = 3/2t^2
20 = (t^2)/2
t^2 = 40
t = 6.32455532
 

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