Physics Question?

Question

A car traveling 45 km/h slows down at a constant 0.50 m/s^2 just by "letting up on the gas." Calculate the distance the car coasts before it stops, the time it takes to stop, and the distance it travels during the first and fifth seconds. Please show/explain all work.

Answer 1

The equation for this is
v(t)=v0-a*t
and
x(t)=v0*t-.5*a*t^2

When the car stops, v(t)=0, solve for t:

0=45/3.6-.25*t^2
t=2*sqrt(45/3.6)

the distance to stop is found by plugging the t into
x(t)=v0*t-.5*a*t^2
x(2*sqrt(5/3.6)=
45/3.6*2*sqrt(45/3.6)-.25*(45/3.6)

to calculate the distance from t=1 to t=5, plug them into x(t)
x(5)-x(1)=
45/3.6*(4)-.25*(25-1)


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