College physics problem.


by chonkyfire
Tags: college, physics
chonkyfire
chonkyfire is offline
#1
Feb27-12, 12:39 AM
P: 1
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.
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th4450
th4450 is offline
#2
Feb27-12, 04:23 AM
P: 38
Draw a v-t graph, area under the graph is the displacement.
But indeed I got -1337.5 m
PeterO
PeterO is offline
#3
Feb27-12, 05:03 AM
HW Helper
P: 2,316
Quote Quote by chonkyfire View Post
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 ; thats -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
th4450 is offline
#4
Feb27-12, 05:34 AM
P: 38

College physics problem.


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


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