Police driving with a velocity of 50 m/s decide to chase a speeder who is 3 km ahead and moving at 55 m/s. The police car accelerates at 2 m/s2. Instantly the speeder becomes aware that he is being chased and starts to accelerate at 1 m/s2. How much time (in s) passes until the police catch the speeder

A body performs a uniformly accelerated rectilinear motion or uniformly varied rectilinear motion when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases its modulus in a uniform way.

The position is calculated by the expression:

x = x0 + v0*t + 1/2*a*t²

where:

x0 is the initial position.
v0 is the initial velocity.
a is the acceleration.
t is the time interval in which the motion is studied.

First, let’s look at the police car’s equations of motion. In this case:

x0= 0
v0= 50 m/s
a= 2 m/s²

So: x = 50 m/s*t + 1/2*2 m/s²*t²

Now for the speeder’s car’s equations of motion you know:

x0= 3 km= 3,000 m
v0= 55 m/s
a= 1 m/s²

So: x = 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

When the police catch the speeder they are both in the same position. So:

50 m/s*t + 1/2*2 m/s²*t²= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t²

Solving:

0= 3,000 m + 55 m/s*t + 1/2*1 m/s²*t² - 50 m/s*t - 1/2*2 m/s²*t²

0= 3,000 + 55 *t + 1/2*t² - 50*t - 1*t²

0= 3,000 + 55 *t - 50*t - 1*t² + 1/2*t²

0= 3,000 + 5*t - 1/2*t²

Applying the quadratic formula:

[tex]x1,x2=\frac{-5+-\sqrt{5^{2}-4*(-\frac{1}{2})*3000 } }{2*(-\frac{1}{2} )}[/tex]

x1= -72.6209

and x2= 82.6209

Since you are calculating the value of a time and it cannot be negative, then the time that passes until the police catch the speeder is 82.6204 seconds.