Question

A bike takes 9.0 seconds to decelerate to a stop over a distance of 30 meters. How fast was the bike traveling before the biker started to decelerate?

Answer 1

You can use the following equation to solve for the initial speed (velocity) of the bike:

`\[v_f^2 = v_i^2 + 2as\]`

Where:

-
`\(v_f\)`is the final velocity (0 m/s because the bike comes to a stop).

-
`\(v_i\)` is the initial velocity (what we want to find).

-
`\(a\)` is the acceleration.

-
`\(s\)` is the distance traveled.

In this case, the bike decelerates, which means the acceleration is negative because it's acting in the opposite direction of motion. So, you have:

`\[0^2 = v_i^2 + 2(-a)s\]`

Now, plug in the values:

`\[0 = v_i^2 + 2(-a)(30\,m)\]`

You know that the time it takes to decelerate to a stop is 9.0 seconds, so you can find the acceleration
`(\(a\))`(\(a\)) using the equation:

`\[0 = v_i^2 + 2(-a)(30\,m)\]`

Since the final velocity
`(\(v_f\))` is 0 m/s and the time
`(\(t\))` is 9.0 seconds, you get:

`\[a = (0 - v_i)/(9.0\,s) = -(v_i)/(9.0\,s)\]`

Now, substitute this expression for acceleration into the previous equation:

`\[0 = v_i^2 + 2\left(-(v_i)/(9.0\,s)\right)(30\,m)\]`

Simplify:

`\[0 = v_i^2 - (60v_i)/(9.0\,s)\]`

Now, you can solve for
`\(v_i\)`:

`\[v_i^2 - (60v_i)/(9.0\,s) = 0\]`



`\[v_i(v_i - 6.67\,m/s) = 0\]`

This equation has two solutions, but the initial velocity cannot be negative (as the bike was moving in the forward direction), so you take the positive solution:

`\[v_i = 6.67\,m/s\]`

So, the bike was traveling at approximately 6.67 meters per second before the biker started to decelerate.