OUNT OF TIME IN SECONDSHow far will the accelerating car overtake the truck?

[Gary King]

Could someone help me to solve this question?

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Question:
At the instant when the traffic light turns green, an automobile starts with a constant acceleration of $$2.00 m/s^2$$. At the same instant, a truck traveling with a constant speed of 9.00 m/s overtakes and passes the automobile. How far beyond the starting point will the automobile overtake the truck?

 

[Andrew Mason]

Could someone help me to solve this question?

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Question:
At the instant when the traffic light turns green, an automobile starts with a constant acceleration of 2.00 m/s^2. At the same instant, a truck traveling with a constant speed of 9.00 m/s overtakes and passes the automobile. How far beyond the starting point will the automobile overtake the truck?

The condition for the car passing the truck is:

$$d_{car} \ge d_{truck}$$

(1)The distance moved by the car as a function of time is: ________
(2)The distance moved by the truck as a function of time is: _______

Set (1) = (2) and you have the expression for the time at which the truck and car have traveled equal distances. Solve for the time.

AM

 

[Gary King]

Sorry to bother you again, but what equations should I use for the car and the truck? I'm having 'one of those days' again, and at the same time I'm having a complete brain fart :)

 

[Andrew Mason]

Plot the velocity as a function of time for the truck and for the car. What gives you the distance covered by the truck and car? (what is velocity in terms of distance and time?).

AM

 

[Gary King]

okay thanks; $$40.5 m$$ is what I got.

 

[Andrew Mason]

okay thanks; $$40.5 m$$ is what I got.

How did you get that? (hint: The area under the graph for the accelerating car is a triangle).

AM

 

[Gary King]

I did:

car

$$2 = \frac {d} {t^2}$$

$$t = \sqrt{d/2}$$

truck

$$d/9 = \sqrt{d/2}$$

$$2d^2 - 81d + 0 = 0$$

Then I just used

$$x = \frac {-b +- \sqrt{b^2 - 4ac}} {2a}$$

To solve for x. I got $$x = 0$$ and $$x=40.5$$. The second one makes sense.

 

[Andrew Mason]

I did:

car

$$2 = \frac {d} {t^2}$$

$$t = \sqrt{d/2}$$

This is not correct. The area under the v-t graph represents the distance traveled by the accelerating car. It is a triangle with base t and height v, so the distance (area) is $\frac{1}{2}vt = \frac{1}{2}(at)t$

So $d_{car} = \frac{1}{2}at^2$

AM