Bike and car acceleration physics

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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)

Homework Equations

all constant acceleration equations

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]

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 traveled an additional 12*20=240 meters.
Thus, you are to solve
320+20t=300+30t, meaning t=2.

 

[m2003]

I would like to point out that the negative sign in your answer indicates that the bike and car are not together at the same time. This is because the bike accelerates at a higher rate than the car, so it reaches its constant speed of 20 m/s before the car reaches its constant speed of 30 m/s. Therefore, the bike will have a head start of 2 seconds before the car catches up to it.

To find the exact moment when they are both at the same displacement, we can set the two displacement equations equal to each other and solve for time.

300 + 30t = 80 + 20t
10t = 220
t = 22 seconds

So, the bike and car will be at the same displacement after 22 seconds, but the bike will be 2 seconds ahead of the car.

I hope this helps clarify the concept of displacement and time in relation to acceleration. Keep up the good work in your studies!