Question

A speeder passes a parked police car at a constant speed of 37 m/s. At that instant, the police car starts from rest with a uniform acceleration of 2.81 m/s2. How much time passes before the speeder is overtaken by the police car?

Answer 1

\[x= \frac{1}{2} a t^2 + vt +x_o\]
where a is acceleration
v is velocity
and xo is the initial distance

let us first consider the initial distance to be zero

the speeder only has velocity of 37 m/s and no acceleration
we can write the distance travelled as

x = 37 t


the police car on the other hand is starting from rest meaning velocity is zero
but has an acceleration of 2.81 m/s^2
we can write the distance travelled as

1/2( 2.81)t^2 = x

now, when the police car finally catches up to the speeder, that would mean that the distance travelled of both the police car and speeder would be the same

so substutite the 2 equations together and solve for t

Answer 2

so the final equation to solve would be 37t = 1/2(2.81)t^2

Answer 3

@completeidiot could you explain how to solve for t and get rid of the t^2?