# A car is traveling at a steady 74 km/h in a 50 km/h zone....

• Mary0524
In summary, a police motorcycle accelerates at a rate of 8.0m/s2 and takes 2.569 seconds to reach a speed of 20.5m/s, traveling a distance of 26m in the process. This can be found using the SUVAT equations for constant acceleration.

## Homework Statement

acceleration =8.0m/s
constant= 74km/h
displacement 50 km/h?

## Homework Equations

How much time elapses before the motorcycle is moving as fast as the car?
How far is the motorcycle from the car when it reaches this speed?

## The Attempt at a Solution

Is there a formula where i can plug my givens in?

It doesn't look like your full question was posted. What is the initial velocity of the motorcycle?

A car is traveling at a steady
74 km/h in a 50 km/h zone. A police motorcycle takes off at the instant the car passes it, accelerating at a steady 8.0 m/s2 .
How much time elapses before the motorcycle is moving as fast as the car?
How far is the motorcycle from the car when it reaches this speed?
I got the first question by doing v= 74*(5/18) which gave me 20.5s
then I looked in my book and got the formula vf=vi +A(T)
vf=20.5
vi=0
a=8.0
so i set 20.5=8.0t and solved for T giving me 2.569

the Second question is what i am confused on

So, it takes the motorcycle 2.569 seconds to accelerate to 20.5m/s.
How far does the motorcycle travel in 2.569 seconds?
You can use the fact that distance traveled is the integral of velocity:
##s(T) = \int_0^T v(t) dt##
and velocity for constant acceleration, as you already showed, is equal to acceleration times time.
##v(t) = at.##

You can keep all the speeds in kmph. Are you familiar with the SUVAT equations?
Edit: I read the acceleration as 8kmph/s by mistake which is actually 8m/s2. So you need to convert km into m.

RUber said:
So, it takes the motorcycle 2.569 seconds to accelerate to 20.5m/s.
How far does the motorcycle travel in 2.569 seconds?
You can use the fact that distance traveled is the integral of velocity:
##s(T) = \int_0^T v(t) dt##
and velocity for constant acceleration, as you already showed, is equal to acceleration times time.
Im taking an algebra based physics class so the calculus scares me. lol is there something a
little more broken down?

For the first part, from SUVAT, you can use v=u+at, which you've already done.
For the second part, you can compute the displacemets of both the vehicles (using SUVAT) and then take the difference between them.

Last edited:
In general, an object starting from rest with constant acceleration will cover a distance of:
##d = \frac12 a T^2 ## over the time from t= 0 to t= T.
If you have an initial velocity, this is changed to
##d = v_0 T + \frac12 a T^2 ## over the time from t= 0 to t= T.
I imagine the SUVAT equations should look similar.

I got 26m and it was correct! thank you so much! :)

• RUber