Question

A student was traveling to see his grandmother who lives 15 miles north oh his home he started from rest and maintained a pace of 4 m/s before stopping at his grandmother's house for lunch which of the following best describes the speed, velocity and acceleration of the student's walk?

Answer 1

Explanation:

Given parameters:

Distance  = 15miles north = 24140.2m

Initial velocity  = 0m/s

Final velocity  = 4m/s

Unknown:

Speed, velocity and acceleration = ?

Solution:

The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.

  Speed  = $\frac{distance }{time}$  

  The speed of the student is 4m/s

Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;

    Velocity  = $\frac{displacement}{time}$  

   The velocity of the student is 4m/s due north

Acceleration is the change in velocity with time;

     To find the acceleration, we use

      v²  = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

     4² = 0² + 2x a x 24140.2

       a  = $\frac{16}{2 x 24140.2}$  = 0.00033m/s²