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

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An automobile accelerates at 2.00 m/s² from a traffic light while a truck travels at a constant speed of 9.00 m/s. To determine when the automobile overtakes the truck, the distances traveled by both vehicles must be equal. The distance for the car is calculated using the formula d_car = 0.5 * a * t², while the truck's distance is d_truck = v * t. After solving the equations, it is determined that the automobile will overtake the truck at a distance of 40.5 meters from the starting point.
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?
 
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Gary King said:
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
 
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 :)
 
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
 
okay thanks; 40.5 m is what I got.
 
Gary King said:
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
 
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.
 
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Gary King said:
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
 

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