Question

A car traveling at 41 feet per second decelerates at a constant 7 feet per second squared. How many feet does the car travel before coming to a complete stop?

Answer 1

\[v ^{2} = u ^{2} + 2as\]

Answer 2

v = initial velocity, u = final velocity, a = acceleration, s = displacement

Answer 3

Can you help show me how to apply this? Thanks

Answer 4

\[0 = 41^{2} + 2 (-7)(s)\]

Answer 5

final velocity is when he stopped so it is 0. initial is 41. acceleration is opposite motion so its negative

Answer 6

thanks :)

Answer 7

\[v = u + at\]
\[v^{2} = u ^{2} + 2as\]
\[s = ut + \frac{ 1 }{ 2 }at ^{2}\]

Answer 8

good equations to remember for kinematics questions

Answer 9

Thanks!