How to Calculate Distance and Speed with Constant Acceleration

[sydneyfranke]

Homework Statement

A car is at a stop light, when the light turns green a truck blows through the intersection at 22 m/s, and remains constant. The car begins to accelerate (assume at the exact same time as the truck goes through intersection) at a rate of 3.55 m/s/s.
A) How far away from its initial point is the car when it passes the truck?

B) How fast is the car going?

Homework Equations

I would think something like v²= V_o² + 2a(x-x_o) because time is not given. But this doesn't seem right.

The Attempt at a Solution

I have tried the above equations. We just learned this stuff today so I'm still not sure what "V_o" actually is.

If someone could give me the "dumb" version of how to figure this one out, I would greatly appreciate it.

 

[NobodySpecial]

You might need to calculate 't' the time the car catches the truck.
Consider s = v₀t + 1/2 at^2

v₀ is the intitial speed, so for the truck it is 22m/s and for the car=0
They obviously meet when 's' and 't' are the same - ie they are at the same place at the same time

 

[sydneyfranke]

okay, but if a of the truck is 0, won't that cancel out the t^2? I'm still a little confused here.

 

[NobodySpecial]

Yes - for the truck it's position at any time 't' is just s = V₀ t
For the car it is s = 1/2 a t² since V₀ is 0

Note there is a second solution at time t=0, where they meet at the start - but you aren't interested in that!

Hint: you could draw a simple time against position graph for each vehicle on a piece of graph paper to give you an idea of the answer.

 

[sydneyfranke]

Okay, so I have t at 12.39 sec., do I plug that into V_ot + 1/2 at² ? That gives me 545 m, but that just doesn't seem reasonable.

 

[NobodySpecial]

22m/s is about 50mph
0-60 in 12.5 secs is reasonable
So you probably aren't an order of magnitude out

Sketch the graph if you aren't sure