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...