Finding Distance with Constant Acceleration

[B18]

Homework Statement

At the instant the traffic light turns green, a car starts with a constant acceleration of 5.00 ft/s^2. At the same instant a truck, traveling with a constant speed of 70.0 ft/s, overtakes and passes the car. How far from the starting point (in feet) will the car overtake the truck?

Homework Equations

Well I know the 4 kinematic equations for motion with constant acceleration.
I know final velocity of the truck=final velocity of the car.

The Attempt at a Solution

To find the time it took for the car to catch the truck i divided 70ft/s by 5.0ft/s^2=14.0s
t=14
Xi=0
Vi=0
a=5.0ft/s^2

Ive got the answers 490ft, 1015ft, 980 ft which are all incorrect. Am I using the wrong value for something here?

I appreciate any help.

 

[NascentOxygen]

Homework Statement

At the instant the traffic light turns green, a car starts with a constant acceleration of 5.00 ft/s^2. At the same instant a truck, traveling with a constant speed of 70.0 ft/s, overtakes and passes the car. How far from the starting point (in feet) will the car overtake the truck?

Hi B18! The question DOES NOT ask how many seconds until the car's speed matches that of the truck!

The question asks: how many feet from their starting point are they seen to be side-by-side? Big difference

 

[B18]

Hi B18! The question DOES NOT ask how many seconds until the car's speed matches that of the truck!

The question asks: how many feet from their starting point are they seen to be side-by-side? Big difference

Hi NascentOxygen. Thanks for the reply. I am sorry I wasn't more specific. I found t=14 s. After plugging in the values I know I wasn't getting the correct answer.

What I mean is the t value I found correct? Did it take the car 14 s to reach the truck??

Wouldn't you need to know the duration of time in this problem??

 

[NascentOxygen]

I haven't worked it out, but I can see that dividing that v by that a will give you the time when the car reaches a speed of 70' per sec. The problem does not require you to determine anything relating to the event of the car's speed matching that of the truck. The problem involves equal distances, not equal speeds.

 

[HalfLostAndSearching]

Thank you.Hi there,

Your approach is correct, but there are a few things you need to consider in order to get the correct answer.

Firstly, you are correct in using the kinematic equations for motion with constant acceleration. However, you need to make sure that you are using the correct equations for the situation. In this case, since you are trying to find distance, you need to use the equation d = vit + 1/2at^2, where d is the distance, vi is the initial velocity, a is the acceleration, and t is the time. This will give you the distance traveled by the car in 14 seconds.

Secondly, you need to consider the initial velocity of the car. The car is starting from rest, so its initial velocity is 0 ft/s. This means that vi = 0 in the equation above.

Thirdly, you need to make sure that you are using the correct final velocity for the car. In this case, the final velocity of the car is not equal to the final velocity of the truck. The car is accelerating, so its final velocity will be greater than 70 ft/s. To find the final velocity of the car, you can use the equation vf = vi + at, where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. In this case, vf = 5 ft/s^2 * 14 s = 70 ft/s. This means that the final velocity of the car is 70 ft/s, not 70 ft/s^2.

Using all of this information, you can now solve for the distance traveled by the car. Plugging in the values into the equation d = vit + 1/2at^2, you get d = 0 + 1/2 * 5 ft/s^2 * (14 s)^2 = 490 ft. This is the correct answer.

I hope this helps! Keep up the good work.