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

To find the initial speed of the bike before deceleration, we can use the equation:

\[v_f^2 = v_i^2 + 2ad\]

where:
\(v_f\) = final velocity (which is 0 m/s since the bike comes to a stop)
\(v_i\) = initial velocity (what we want to find)
\(a\) = acceleration (since the bike is decelerating, the acceleration is negative)
\(d\) = distance traveled (30 meters)

Plugging in the given values, we get:

\(0^2 = v_i^2 + 2(-a)(30)\)

Simplifying, we have:

\(0 = v_i^2 - 60a\)

Since we know that the time it takes to decelerate to a stop is 9.0 seconds, we can use the equation:

\(a = \frac{{\Delta v}}{{\Delta t}}\)

to find the acceleration. In this case, the change in velocity is \(0 - v_i = -v_i\) and the change in time is 9.0 seconds. Substituting these values into the equation, we get:

\(a = \frac{{-v_i}}{{9.0}}\)

Plugging this value of acceleration back into the initial equation, we have:

\(0 = v_i^2 - 60 \left(\frac{{-v_i}}{{9.0}}\right)\)

Simplifying further:

\(0 = v_i^2 + \frac{{-60v_i}}{{9.0}}\)

Multiplying through by 9.0 to remove the fraction:

\(0 = 9.0v_i^2 - 60v_i\)

Rearranging the equation:

\(9.0v_i^2 - 60v_i = 0\)

Factoring out \(v_i\):

\(v_i(9.0v_i - 60) = 0\)

Setting each factor equal to zero and solving for \(v_i\):

\(v_i = 0\) (which is not possible since the bike was definitely moving)
\(v_i = \frac{{60}}{{9.0}}\)

Using a calculator, we can find that \(v_i \approx 6.67\) m/s.

Therefore, the bike was traveling at approximately 6.67 meters per second (m/s) before the biker started to decelerate.