# At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.1 m/s^2. At the same instant a truck, traveling with a constant speed of 9.2 m/s, overtakes and passes the automobile ?

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Time needed:

Speed of car:

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**FULL QUESTION**

*At the instant the traffic light turns green, an automobile starts with a constant acceleration a of* *At the same instant a truck, traveling with a constant speed of #"9.2 m/s"#, overtakes and passes the automobile.*

*(a) How far beyond the traffic signal will the automobile overtake the truck?*

*(b) How fast will the automobile be traveling at that instant?*

This one is pretty straightforward. You know that at

At the same time, a truck moving at a speed of

The idea here is that the **distance** they travel until the car overtakes the truck is **the same** for both the car, and the truck.

The same can be said for the **time** that passes until the two meet again.

So, if the car takes a time

#d = v * t -># for the truck

and

#d = underbrace(v_0)_(color(blue)(=0)) * t + 1/2 * a * t^2 -># for the car

This will get you

#v * color(red)(cancel(color(black)(t))) = 1/2 * a * t^color(red)(cancel(color(black)(2))) implies t = (2 * v)/a#

#t = (2 * 9.2color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))/(2.1color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2)))) = "8.76 s"#

This means that the distance is

#d = v * t = 9.2"m"/color(red)(cancel(color(black)("s"))) * 8.76color(red)(cancel(color(black)("s"))) = color(green)("80.6 m")#

The speed of the car is

#v_"car" = a * t = 2.1"m"/"s"^color(red)(cancel(color(black)(2))) * 8.76color(red)(cancel(color(black)("s"))) = color(green)("18.4 m/s")#

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