Time to a driver catch another

[duplaimp]

Homework Statement

Two drivers met at a station and they discover that both have the same destination.
The driver A has a car which goes from 0 to 100km/h in 16s and driver B has a car which goes from 0 to 100km/h in 5s. The maximum velocity of each is 100km/h
How much time after driver A starts driving should driver B start driving to be sure that he catches driver A?

Homework Equations

Should I use constant acceleration equations?

The Attempt at a Solution

I tried the following:
100km/h = 27.8m/s

Driver A
a = 27.8/16 = 1.73m/s^2

Driver B
a = 27.8/5 = 5.56m/s^2

Now I can't continue because I don't know if I can use constant acceleration equations

 

[tiny-tim]

The maximum velocity of each is 100km/h
…
Now I can't continue because I don't know if I can use constant acceleration equations

yes of course

start by finding how long it takes the first one to reach 100 km/h (after which it stays at that speed while the other one catches up)

 

[duplaimp]

To reach 100km/h it takes 16s and the car B takes 5s
But I still can't figure how to get how long after driver A starts driving should driver B start to reach it

 

[tiny-tim]

write an equation for the first car after it reaches 100 km/hr, and an equation for the second car while it is still accelerating

start one car at t = 0, and the other at t = t_o

 

[duplaimp]

$x-x_{0}=\frac{1}{2}(1.73)*16^{2}$
$x-x_{0}=\frac{1}{2}(5.56)*t^{2}$

Something like this?

 

[tiny-tim]

$x-x_{0}=\frac{1}{2}(5.56)*t^{2}$

yes

$x-x_{0}=\frac{1}{2}(1.73)*16^{2}$

no …

write an equation for the first car after it reaches 100 km/hr

also …

start one car at t = 0, and the other at t = to

 

[duplaimp]

$27.8^{2}=2(1.73)(x-x_{0})$
$x-x_{0}=\frac{1}{2}(5.56)*t^{2}$

This?

 

[tiny-tim]

??

after it reaches 100 km/hr, it has constant speed

 

[duplaimp]

Yes, but with this the acceleration will be 0, right?

 

[tiny-tim]

Yes, but with this the acceleration will be 0, right?

yes