How long does it take for the car to pass the truck?

In summary: Here is the complete equation:x(t) = Xo + Vo*t + 1/2*a*t^2You need to use that equation to write the equations for each vehicle. Can you do that?x(t) = Xo + Vo*t + 1/2*a*t^2For the first car:x(t) = 0 + 25*t + 1/2*1*t^2x(t) = 25t + 0.5t^2For the truck:x(t) = 10 + 25*t + 0*t^2x(t) = 25t +
  • #1
physicsguy69
3
0

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?
 
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  • #2
physicsguy69 said:

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
physicsguy69 said:
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
physicsguy69 said:
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...
 

1. What is constant acceleration?

Constant acceleration is the rate of change of velocity over time that remains the same. This means that the velocity of an object increases or decreases by the same amount every second.

2. How is constant acceleration calculated?

Constant acceleration is calculated by dividing the change in velocity by the change in time. The formula for constant acceleration is a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What are some examples of motion with constant acceleration?

Some examples of motion with constant acceleration include a car accelerating from a stop, a ball dropped from a height, and a roller coaster going down a hill. In each of these cases, the velocity changes by the same amount every second.

4. How does constant acceleration differ from uniform motion?

Constant acceleration and uniform motion are similar in that the velocity changes by the same amount every second. However, in constant acceleration, the rate of change of velocity is constant, while in uniform motion, the velocity remains the same over time.

5. What is the significance of constant acceleration in physics?

Constant acceleration is a fundamental concept in physics and is used to describe the motion of objects under the influence of a constant force. It helps us understand how objects move and how forces affect their motion.

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