# A race car speeds up. What is it's velocity and its acceleration at 0s, and 0m, at 0.25s and 0.11m, at 0.50s and 0.46m, at 0.75s and 1.06m?

## Expert Answers 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`

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Further Reading

Approved by eNotes Editorial Team 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

Approved by eNotes Editorial Team

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