How far beyond its starting point does the car overtake the truck?

Question

At the instant the traffic light turns green, a car that has been waiting at an intersection starts ahead with a constant acceleration of 3.2 m/s2 . At the same instant a truck, traveling with a constant speed of 20 m/s overtakes and passes the car at the same intersection. The car eventually overtakes the truck. Draw a motion diagram for both the car and the truck.

Answer 1

We know that the car was waiting at the light, so its initial velocity is 0 and we are told that it then accelerates at a constant rate of 3.2 m/s^2. Using the kinematic equation that connects distance, time, initial velocity, and acceleration, d=v(init)*t+1/2*a*t^2, we substitute d=0t+1/2(3.2*t^2)=1.6t^2.

The truck passed the car at the same moment the car began accelerating, while traveling at 20 m/s. Relating distance to velocity, we know that d=vt, and substituting we say that d=20t.

Now, we know that when the car catches up to the truck it they will be the same distance from the traffic light, so we can say that d(car)=d(truck), or 1.6t^2=20 t.

Using algebra, we can divide both sides by t to say that 1.6t=20.

And now we can divide both sides by 1.6 to say that t=20/1.6=12.5

We have now determined that the car will catch up to the truck 12.5 seconds after the moment when the car begins to accelerate as the truck passes it.

Knowing that the last person we want to catch our mistake is the teacher, we can try to catch any mistakes first. To do this, we can substitute our solution time of 12.5seconds back into the two original equations to check that they car and truck have indeed traveled the same distance.

For the car, d = 0*12.5 + 1/2 * 3.2 (12.5^2) = 0 + 1.6 * 156.25 = 250 meters.

For the truck, d = 20(12.5) = 250 meters.

We have now checked our work and found that the car catches up to the truck 250 meters and 12.5 seconds from the light.

Answer 2

Start by listing the known information for the car and the truck, then use this information to write equations for the distance each travels.

For the car, initial speed = 0 and acceleration = 3.2 m/s^2, so d = v initial t + 1/2 a t^2 yields d = 1.6 t^2.

For the truck, v is a constant 20 m/s, so d = 20 t.

For the car to overtake the truck, it must travel the same distance in the same time. Set the two distance expressions equal to one another, then solve for time. Letting d car = d truck, we get 1.6 t^2 = 20 t. Solving for t gives us t = 12.5 s. This is how long it takes the car to catch the truck.

Now substitute t = 12.5 s into each of the distance expressions. for the car, d = 1.6 (12.5^2) = 250 m. For the truck, d = 20(12.5) = 250 m. Happily, they traveled the same distance, so the car has in fact overtaken the truck, at a point 250 m beyond the start.