# Equations of Motion – Constant Acceleration Intercept Example Problem 1

Motion in a straight line with constant acceleration is the simplest form of accelerated motion. Since acceleration is constant, the velocity changes at the same rate as time progresses. Position will change as the square of the time progressing. There are three basic formulas that will help with most homework problems dealing with motion under constant acceleration.

(1) x = x0 + v0t + ½at2

(2) v = v0 + at

(3) v2 = v02 + 2a(x – x0)

where
x is the distance travelled
x0 is the initial starting point
v is the velocity
v0 is the initial velocity
a is the acceleration
t is the time

This worked example problem shows how to use the equations of motion to find the time it takes a constantly accelerating body to intercept another body moving at a constant velocity.

Example Problem:
A speeding motorist travelling at 120 km/hr passes a stopped police car. The police car immediately begins to chase the speeder, accelerating at a constant 2.5 m/s2.
(a) How long does it take for the police car to intercept the speeder?
(b) How far did the police car travel before catching up to the speeder?
(c) How fast was the police car travelling when it intercepts the speeder?

Solution:
This illustration shows the conditions of the vehicles at the beginning of the problem and the time when the police car intercepts the speeder.

Part a) How long does it take for the police car to intercept the speeder?

First, let’s look at the police car’s equations of motion.

xPI = x0P + V0Pt + ½at2

since the police car starts at 0 and at rest, v0P=0 then

xP = ½at2

vPI = v0P + at

vPI = at

Now for the speeder’s car’s equations of motion.

xS = x0S + V0St + ½at2

x0 = 0 and the speeder is not accelerating, a = 0, therefore

xS = v0St

vS = v0S + at

vS = V0S

vS = 120 km/hr

Convert to m/s since our acceleration is in m/s2 and it probably won’t take hours for the police car to catch up.

cancel out the units and we are left with m/s

multiply out to get

vS = 33.3 m/s

The two vehicles were in the same position at the very beginning of the chase at x = 0. We need to find where that happens again. This will happen when xPI = xS.

From above:

xPI = ½at2

xS = v0St

Since xPI = xS, we can set these two equations equal to each other.

½at2 = v0St

This quadratic equation has two solutions. The first solution for t is at t = 0 seconds. To find the second, we can divide both sides by t.

½at = v0S

Solve for t

plug in the values for a = 2.5 m/s2 and v0S = 33.3 m/s and get

t = 26.6 s

It takes 26.6 seconds for the police car to catch up and intercept the speeder.

Part b) How far did the police car travel before catching up to the speeder?
Now that we know the time, we can find the distance. From the police car’s position equation above:

xPI = ½at2
xPI = ½(2.5 m/s2)(26.6)2
xPI = 888.4 m

The police car travelled 888.4 m before it intercepted the speeder.

Part c) How fast was the police car travelling when it intercepts the speeder?
Again, using the time and the police car’s velocity equation from above:

vPI = at
vPI = (2.5 m/s2)(26.6 s)
vPI = 66.7 m/s

The police car was travelling at 66.7 m/s when it intercepted the speeder. If you convert it to km/hr, the speed of the police car is 239.9 km/hr. Talk about speeding vehicles.

For another example of constant acceleration motion, check out Equations of Motion – Constant Acceleration Example Problem. This problem shows how to find details about the position, velocity and acceleration of a breaking vehicle.