A man speeding at 40m/s decides to outrun the cops and starts to accelerate at a rate of 2.5m/s² for 12 seconds. How far did he travel in this time?

The distance traveled by the man during the acceleration is 1186.32 meters

Explanation:

The initial velocity of the man = 40 m/s

The added acceleration rate of the man to outrun = 2.5 m/s²

The time of acceleration = 12 seconds

The relevant kinematic equation of motion is given as follows;

s = u·t + 1/2·a·t²

Where;

u = The initial velocity of the car the man is driving = 40 m/s

t = The time of travel during acceleration of the car the man is driving = 12 seconds

a = The acceleration by the man to outrun the cops = 2.5 m/s²

s = The distance traveled by the man during the acceleration

Substituting, gives;

s = 40 × 12 + 1/2 × 9.81 × 12² =  1186.32

The distance traveled by the man during the acceleration = s = 1186.32 m.

what was the answer?

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west