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?
Answers
GPT 3.5
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.
\[v_f^2 = v_i^2 + 2ad\]
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.