Chase problem involving acceleration and velocity

[patton_223]

Homework Statement

Car A accelerates from rest at 0.75m/s^2 for 1 minute. After this Car A cruises at a constant speed. Car B accelerates from rest at the same time at 0.1m/s^2. When will car B catch up with Car A? What distance have they traveled and what is the final speed of each car?

Homework Equations

d= ViT + 1/2at^2
Vf= Vi + aT

The Attempt at a Solution

i found the speed of car A which is 22.5 m/s and Car B which is 45 m/s

and i did the calculations and i got 450 sec for the time for Car B to catch up with car A

but when i tried to find the distance i got really odd numbers, i got 10125m

because i would plug in 450 sec into (1/2)(0.1)(450^2) and got that

i just want to know if i did this right, and i could use some help if this is wrong

 

[Delphi51]

Welcome to PF, Patton!
How did you find the speed of car A after 60 seconds?
I used V = Vi + a*t = 0 + .75*60 and got 45 m/s.
No velocity for car B can be found because we have no time for it . . so far.

The big picture on a chase problem is to write
car A distance = car B distance
Then you just need a complete formula for each distance and solve the equation for time. I'm pretty sure the meet will be after the initial 60 seconds so you only need formulas valid after that time. If you have a go at the two formulas, we'll be glad to help you make sure they are correct.

I checked your time of 450. At that time car A has gone the initial .5*.75*60² = 1350 m plus the d = v*t = 45*(450-60) = 17550 m for a total of 17900 m. Looks like car B still has a long way to catch up at time 450!

 

[patton_223]

Welcome to PF, Patton!
How did you find the speed of car A after 60 seconds?
I used V = Vi + a*t = 0 + .75*60 and got 45 m/s.
No velocity for car B can be found because we have no time for it . . so far.

The big picture on a chase problem is to write
car A distance = car B distance
Then you just need a complete formula for each distance and solve the equation for time. I'm pretty sure the meet will be after the initial 60 seconds so you only need formulas valid after that time. If you have a go at the two formulas, we'll be glad to help you make sure they are correct.

I checked your time of 450. At that time car A has gone the initial .5*.75*60² = 1350 m plus the d = v*t = 45*(450-60) = 17550 m for a total of 17900 m. Looks like car B still has a long way to catch up at time 450!

I found the speed of car a by using d=1/2aT^2 i used the acceleration and 60sec and found the distance to be 1350m and then divided that by 60sec to find the velocity of 22.5

but it's the distance that's throwing me off

 

[Delphi51]

I found the speed of car a by using d=1/2aT^2 i used the acceleration and 60sec and found the distance to be 1350m and then divided that by 60sec to find the velocity of 22.5

That 1350 m for the first 60 seconds is correct. But you used d = vt to find the velocity and d=vt does not apply to this portion of the motion because it is accelerated. You must use V = Vi + at to find the velocity at time 60.