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Distance/Time Problem
Problem: A car traveling 55 km/h slows down at a constant 0.50 m/s^2. Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and fifth seconds.
v(initial) = velocity(initial) = 55km/h = 15.28m/s
v = velocity
a = acceleration = -0.50m/s^2
For (a), I used the equation v^2 = v(initial)^2 + 2a(x - x(initial)) to get the resulting displacement of 230 meters.
For (b), I found the time using the equation v = v(initial) + at, which resulted in 31 seconds.
However, I really have no idea how to approach (c). Any help would be greatly appreciated.
v(initial) = velocity(initial) = 55km/h = 15.28m/s
v = velocity
a = acceleration = -0.50m/s^2
For (a), I used the equation v^2 = v(initial)^2 + 2a(x - x(initial)) to get the resulting displacement of 230 meters.
For (b), I found the time using the equation v = v(initial) + at, which resulted in 31 seconds.
However, I really have no idea how to approach (c). Any help would be greatly appreciated.
Find where it is after 1s
Find where it is after 4s
Find where it is after 5s
Subtract.
Find where it is after 4s
Find where it is after 5s
Subtract.
for c) you have u, t and a
for part 2, a is the same, t is still a second but you need a new u which you can work out with v=u+at and plugging in the figures and 4 seconds
for part 2, a is the same, t is still a second but you need a new u which you can work out with v=u+at and plugging in the figures and 4 seconds
Thanks for the help!
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