Calculate Final Position of a Car Using Acceleration on a Level Road

[chonkyfire]

1.
Starting from the origin, a car accelerates on a level road from rest at a(A)=-2.00 m/s^2 until T(1)=20.0s. The velocity is then held constant until T(2)=40.0s, and then the car accelerates at a(B)=5.00m/s^2 until T(3)=50.0s.

question: What is the final position of the car?
2. delta X
3. got -400m. but the answer should be -1350m.

 

[th4450]

Draw a v-t graph, area under the graph is the displacement.
But indeed I got -1337.5 m

 

[PeterO]

1.
Starting from the origin, a car accelerates on a level road from rest at a(A)=-2.00 m/s^2 until T(1)=20.0s. The velocity is then held constant until T(2)=40.0s, and then the car accelerates at a(B)=5.00m/s^2 until T(3)=50.0s.

question: What is the final position of the car?



2. delta X



3. got -400m. but the answer should be -1350m.

20 seconds of acceleration achieves -40 m/s; so avergae vel -20.
-20 x 20 = -400 m while accelerating

A further 20 seconds at -40m/s covers a further -800m - that's -1200 so far.

We now accelerate at +5, so it takes 8 seconds to stop.
Average velosity -20, time 8 sec means a further -160m ; that's -1360m so far.

The acceleration continues for a further 2 seconds - reaching a velocity of +10 m/s.
Thats an averag of +5 m/s for 2 seconds, so a displacement of + 10m

-1360 + 10 = -1350

SO how did you get your -400 m ? [or -1337.5 for that matter]

 

[th4450]

Oh yes, should be -1350 m, cal a wrong time interval for deceleration...

 

[asteriatic]

I would like to clarify that the final position of the car will depend on several factors such as the initial position, the direction of motion, and the total time elapsed. In this scenario, the car starts from the origin and moves in one direction only, so we can assume that the final position will also be in the same direction.

To calculate the final position of the car, we need to use the equation: d = v0t + 1/2at^2, where d is the final position, v0 is the initial velocity, a is the acceleration, and t is the time elapsed. We will also need to consider the different time intervals given in the question.

1. From t = 0s to t = 20s, the car accelerates at a(A) = -2.00 m/s^2. The initial velocity, v0, is 0 m/s as the car starts from rest. Using the above equation, we can calculate the final position as:

d1 = (0)(20) + 1/2(-2.00)(20)^2 = -400m

2. From t = 20s to t = 40s, the car moves with a constant velocity, so the acceleration, a, is 0 m/s^2. Using the same equation, we can calculate the final position as:

d2 = (0)(20) + 1/2(0)(20)^2 = 0m

3. From t = 40s to t = 50s, the car accelerates at a(B) = 5.00 m/s^2. The initial velocity, v0, is the constant velocity of 20 m/s from the previous time interval. Using the same equation, we can calculate the final position as:

d3 = (20)(10) + 1/2(5.00)(10)^2 = 1350m

Therefore, the final position of the car will be the sum of all these intervals:

d = d1 + d2 + d3 = -400 + 0 + 1350 = 950m

The negative sign indicates that the car has moved in the negative direction, which is consistent with the given information that the car starts from the origin.

In conclusion, the final position of the car is 950m in the negative direction. It is important to note that this calculation