The distance x of a skier from a fixed point is measured in

The distance x of a skier from a fixed point is measured (in meters) at intervals of 0.25 second. The data obtained is Use central difference to calculate the skier\'s velocity at t = 0.25, and t = 0.375. Calculate the skier\'s acceleration at time t = 0.75.

Solution

4.1) Using the central difference

x(0) = 0, x(0.5) = 7.2

Velocity at t=0.25s is

=> [x(0.5) - x(0)]/(0.5-0)

=> (7.2-0)/0.5 = 14.4 m/s

at x=0.375, we need to take the values of x(0.75) and x(0)

Velocity at t=0.375 is

Velocity = [x(0.75) - x(0)]/(0.75-0) = (12.7-0)/0.75 = 12.7 * 4/3 = 50.8/3 = 16.93 m/s

4.2)

For calculating the skier\'s acceleration

v(0.75) = [x(1) - x(0.5)]/(1 - 0.75) = (20.3-12.7)/0.5 = 15.2 m/s

Distance at 0.75m is given

Now using the equation

s = ut + 1/2 * at^2

12.7 = 15 * 0.75 + 1/2 * a * (9/16)

1.45 = a * 9/32

a = 1.45 * 32/9 = 5.15 m/s^2