A motorcyclist applies brakes with a retardation of 3 ms ^-2 and comes to a stop in 6 seconds. What was the speed at which he was driving ? ( I need the step by step process and make sure your answer is 18 ms^ -1..Show the application of the formulae used. )
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Answer:
a = 3m/s^2
t = 6 seconds
vf = 0 m/s
vi = ??
vf = vi + at
0 = x + (3)(6)
x = -18
but the negative wont be taken into consideration because it is hitting the breaks therefore the answer is just 18 m/s :)
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Answer:
it's in the attachment