Question

A cyclist is riding along at a speed of 20.7 when she decides to apply the brakes which gave a deceleration applied was a rate of -3.4 m/s2 over the span of 7.8 s. What distance does she travel over that period of time.

Answer 1

The distance is 58.03 m

Step-by-step explanation:

Constant Acceleration Motion

It occurs when the velocity of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, the following relation applies:

`v_f=v_o+at`

The distance traveled by the object is given by:

`\displaystyle x=v_o.t+(a.t^2)/(2)`

The conditions of the problem state the cyclist has an initial speed of v0=20.7 m/s during t=7.8 seconds and acceleration of -3.4 m/s^2.

The final speed is:

`v_f=20.7+(-3.4)\cdot 7.8`

`v_f=20.7-26.52`

`v_f=-5.82\ m/s`

Note the cyclist has stopped and come back because his speed is negative. Now calculate the distance:

`\displaystyle x=20.7\cdot 7.8+((-3.4)\cdot 7.8^2)/(2)`

`\displaystyle x=161.46-103.43`

x=58.03 m