Constant Acceleration

  • #1
sydneyfranke
73
0

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= Vo2 + 2a(x-xo) 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 "Vo" actually is.

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

Answers and Replies

  • #2
NobodySpecial
380
3
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
 
  • #3
sydneyfranke
73
0
okay, but if a of the truck is 0, won't that cancel out the t^2? I'm still a little confused here.
 
  • #4
NobodySpecial
380
3
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.
 
  • #5
sydneyfranke
73
0
Okay, so I have t at 12.39 sec., do I plug that into Vot + 1/2 at2 ? That gives me 545 m, but that just doesn't seem reasonable.
 
  • #6
NobodySpecial
380
3
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
 

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