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College physics problem. 
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#1
Feb2712, 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. 


#2
Feb2712, 04:23 AM

P: 38

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


#3
Feb2712, 05:03 AM

HW Helper
P: 2,316

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] 


#4
Feb2712, 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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