Let us begin by drawing a graph of the distance versus time, with the data provided in the problem. See figure below. The figure also contains the **perfect fit** of the given data with the curve

`d(t) =d_0 + v_0*t + a*t^2/2`

with the parameter values

`d_0 =0 m ` , `v_0 =0 m/s` , `a =2*1.96 =3.92 m/s^2`

This is the equation of space in an **uniform accelerated** motion. Therefore the acceleration is the same at any given moment in time

(0 seconds, 0.25 seconds, 0.5 seconds and 0.75 seconds)

**a =3.92 m/s^2 **

Now we know that in an uniform accelerated motion the law of speed variation with time is

`v(t) =v_0 + a*t`

which means that

`v(0) =0 +3.92*0 =0 m/s`

`v(0.25) = 0.3.92*0.25 =0.98 m/s`

`v(0.5) =0+ 3.92*0.5 =1.96 m/s`

`v(0.75) =0 +3.92*0.75 =2.94 m/s`

**Further Reading**

A race car speeds up. Assume the car starts at 0 m/s and the acceleration of the car in each of the intervals is constant and equal to the acceleration derived at the end of the interval.

The car has traveled 0.11 m in 0.25 s.

0*0.15 + (1/2)*a*0.25^2 = 0.11

=> a = 0.22/0.25^2 = 3.52 m/s^2

The speed of the car at this point is 3.52*0.25 = 0.88 m/s

At 0.5 s the car has traveled 0.46 m.

0.88*(0.5 - 0.25) + (1/2)*a*0.25^2 = 0.35

=> a = 4.16 m/s^2

The speed of the car at this point is 0.88 + 4.16*0.25 = 1.92 m/s

At 0.75 s the car has traveled 1.06 m.

1.92*0.25 + (1/2)*a*0.25^2 = 0.6

=> a = 3.6 m/s^2

The speed of the car at this point is 1.92 + 0.25*3.6 = 2.67 m/s

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