Kinematics truck moving at 30m/s suddenly notices a moose 70m

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The discussion revolves around a physics problem involving a truck traveling at 30 m/s that must stop to avoid hitting a moose 70 m ahead. The driver has a reaction time of 0.5 seconds and a maximum deceleration of 8 m/s². Calculations show that the total stopping time is 4.25 seconds, leading to a stopping distance of 55.25 m. However, the book states that the truck needs an additional 1.3 m to stop, indicating a discrepancy in the calculations. Participants suggest treating the reaction time and braking distance as separate segments to accurately determine the distance covered before braking begins.
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Homework Statement


The driver of a truck moving at 30m/s suddenly notices a moose 70m straight ahead. If the driver's reaction time is 0.5s and the maximum deceleration is 8m/s^2, can he avoid hitting the moose without steering to one side?


The Attempt at a Solution


First I found out how long it would take the driver to stop.
v = u + at
0 = 30 -8t
t = 3.75

I then added the drivers reaction time of 0.5s so we land up with 4.25s

Then I tries to fid out the distance covered by using:
s = ut + 1/2at^2
s = 30(4.25) + 1/2(-8)(4.25)^2
s = 127.5 - 72.25
s = 55.25m

It would seem that the truck stops before 70m but the answer given in the book says no he doesn't stop in time and needs a further 1.3m.

Please help.
 
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monkfishkev said:
I then added the drivers reaction time of 0.5s so we land up with 4.25s
Instead of adding the times together, treat the two segments of motion separately. Hint: How far does he get before he steps on the brakes?
 

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