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Motion, constant acceleration

  • #1

Homework Statement



a car is behind a truck going 25m/s on the highway. the car's river looks for a chance to pass, guessing that his car can accelerate at 1.0m/s^2. he gauges that he has to cover the 20 meters length of the truck 10 meters clear room at the rear of the truck and 10 meters more at the front of it. in the oncoming lane, he sees a car approaching, probably also traveling at 25m/s. he estimates that the car is about 400m away

Homework Equations



v^2=Vo^2+2ax
v=Vo+at
d=rt

The Attempt at a Solution



v^2=(2)(1)(40) V=8.9

8.9+25=33.9m/s
1149.2=625+2(1)(x) x=262.1m
it takes car one 262.1meters to pass the truck and this process takes 8.9 seconds

d=rt
d=(8.9sec)(25m/s) d=222.5m

400 - 262.1 ≠ 222.5
400 - 262.1 =137.9

final answer = no

is there anything wrong here?
 

Answers and Replies

  • #2
berkeman
Mentor
57,509
7,531

Homework Statement



a car is behind a truck going 25m/s on the highway. the car's river looks for a chance to pass, guessing that his car can accelerate at 1.0m/s^2. he gauges that he has to cover the 20 meters length of the truck 10 meters clear room at the rear of the truck and 10 meters more at the front of it. in the oncoming lane, he sees a car approaching, probably also traveling at 25m/s. he estimates that the car is about 400m away

Homework Equations



v^2=Vo^2+2ax
v=Vo+at
d=rt

The Attempt at a Solution



v^2=(2)(1)(40) V=8.9

8.9+25=33.9m/s
1149.2=625+2(1)(x) x=262.1m
it takes car one 262.1meters to pass the truck and this process takes 8.9 seconds

d=rt
d=(8.9sec)(25m/s) d=222.5m

400 - 262.1 ≠ 222.5
400 - 262.1 =137.9

final answer = no

is there anything wrong here?
Welcome to the PF.

It looks like you are taking a reasonable approach, but it is a little hard to follow your reasoning. If you could label what you are doing in each step, that would help.

Also, I would (personally) approach it a little differently. I would write 3 equations for the positions of each of the 3 vehicles, something like

Xc1(t) =
Xc2(t) =
Xt(t) =

Use the initial conditions you are given in the problem (like initial positions and speeds), and then I would solve for the time t when the two cars pass each other. And given that time t, would the first car be the 10m past the truck or not. Doing it that way is more intuitive to me, but different folks will approach problems in different ways.

Still, you could use my approach to check your answer...
 
  • #3
what would your equations be?
 
  • #4
berkeman
Mentor
57,509
7,531
what would your equations be?
What is the basic kinematic equation for situations where there is a constant accelertaion (like questions involving gravitational acceleration, or this problem with the constant car acceleration)?

The basic equation is for the distance as a function of time x(t), in terms of the initial position, initial velocity, acceleration, and time. Can you write that general equation? And then use it 3 times, using the appropriate Xo, Vo, etc. for each of the 3 vehicles...
 
  • #5
x=1/2at^2 but the truck and the other car have a constant velocity
 
  • #6
berkeman
Mentor
57,509
7,531
x=1/2at^2 but the truck and the other car have a constant velocity
That's not the whole equation. You left off the Xo and Vo terms...
 

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