Problem with distance in relation to acceleration.

[BeeGeeks]

Homework Statement
A 1200 kg car is at a red light on a horizontal road. When the road turns to green, (t0) he starts accelerating and reaches 14 m/s in 7 seconds. He then continues at a constant speed for 10 seconds, then brakes with a -3.5 m/s^2 acceleration.

All accelerations are assumed to be constant.

a) Calculate the amount of time he spends decelerating. (What I found through graphing : 4 seconds)

b) Graph the speed in relation to time starting with t0. (Done successfully)

c) Calculate the average speed in both the acceleration and deceleration phase. (7 m/s)<- this seems weird to me)

d) Calculate the total distance between the start of the acceleration until the final stop.

I'm having trouble with point d).

The problem

I don't understand how we can calculate a distance while the object is under acceleration. I can't wrap my head around it. Is there an equation I don't know about? Or am I overcomplicating things and it's just ( for this problem) an additional 2 meters per second during the acceleration phase, then 10 seconds of adding 14 m per second and then decreasing the number you add to your total by 3 every second starting by 14 during your deceleration? is it that obvious?

Thanks in advance!

 

[Doc Al]

I don't understand how we can calculate a distance while the object is under acceleration.

Why not? It's moving isn't it? What's the average speed during each segment of the motion? Use that to find the distance traveled.

 

[BeeGeeks]

Why not? It's moving isn't it? What's the average speed during each segment of the motion? Use that to find the distance traveled.

Thanks!

 

[CWatters]

Well worth memorising the four/five basic equations of motion. See "SUVAT equations" down here..

http://en.wikipedia.org/wiki/Equations_of_motion

 

[Nickie]

Calculating distance while an object is under acceleration can be tricky, but it is possible using a few key equations. In this case, we can use the equation d = vt + 1/2at^2 to calculate the distance traveled during each phase of the car's motion.

During the acceleration phase, the car's initial velocity is 0 m/s and its final velocity is 14 m/s. We also know that the acceleration is 2 m/s^2 (since the car reaches 14 m/s in 7 seconds). Plugging these values into the equation, we get d = (14 m/s)(7 s) + 1/2(2 m/s^2)(7 s)^2 = 49 m + 49 m = 98 m.

During the constant speed phase, the car's velocity remains at 14 m/s for 10 seconds. Using the equation d = vt, we get d = (14 m/s)(10 s) = 140 m.

During the deceleration phase, the car's initial velocity is 14 m/s and its final velocity is 0 m/s. We also know that the acceleration is -3.5 m/s^2. Plugging these values into the equation, we get d = (14 m/s)t + 1/2(-3.5 m/s^2)t^2 = 14t - 1.75t^2.

Now, we can use the value we found for the time spent decelerating (4 seconds) to calculate the distance traveled during this phase. Plugging in t = 4, we get d = (14 m/s)(4 s) - 1.75(4 s)^2 = 56 m - 28 m = 28 m.

Therefore, the total distance traveled by the car is 98 m + 140 m + 28 m = 266 m.

I hope this helps to clarify the process of calculating distance while an object is under acceleration. It's important to remember to use the appropriate equations for each phase of motion and to pay attention to the units being used. Keep practicing and you'll become more comfortable with these types of problems in no time!