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Bike and car acceleration physics

  • Thread starter fffff
  • Start date
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1. Homework Statement
a car accelerates at 1.5ms^-2 from 0 ms^-1 to 30 ms^-1. It then travels at a constant speed of 30 ms^-1.

2.A bike accelerates at 2.5ms^-2 from 0ms^-1 to 20 ms^-1 and then travels at a constant speed of 20 ms^-1

find the instant where there together.
(Both have the same displacement)

2. Homework Equations
all constant acceleration equations


3. The Attempt at a Solution

for part one, i can represent the distance as 300+30t.
t=(30-0)/1.5=20s

S=1/2(20)(30) + 30t= 300 +30t

for part 2.

t=(20-0)/2.5=8s

S=1/2(8)(20) + 20t= 80 + 20t


80+20t=300+30t
-220=10t
t=-22s

why is my answer coming in negative. My answer -22 is right but its negative.
 

arildno

Science Advisor
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This is totally wrong!
Your flaw lies in equating the times when both vehicles start moving with constant speed.
Now, if we let t=0 represent the time when the CAR reaches constant speed, then during the twelve seconds from the BIKE reached constant speed to t=0, the bike has travelled an additional 12*20=240 meters.
Thus, you are to solve
320+20t=300+30t, meaning t=2.
 

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