A motor cyclist travelling at 12m/s decelerate at 3m/s(squared)? How far does he travel while coming to rest?
V^2 = Vo^2 + 2a*d
V = 0
Vo = 12 m/s
a = -3 m/s^2
Solve for d.
How far do you travel when you acclerate at a rate of 2.3m/s2 while initially travelling at 12m/s and ending at a velocity of 19m/s?
To find the distance travelled while decelerating, we can use the kinematic equation:
\(v^2 = u^2 + 2as\)
Where:
\(v\) = final velocity (0 m/s, since the motorcyclist comes to rest)
\(u\) = initial velocity (12 m/s)
\(a\) = acceleration/deceleration (-3 m/s^2, considering the negative sign indicates deceleration)
\(s\) = distance travelled
Rearranging the equation to solve for \(s\), we have:
\(s = \frac{v^2 - u^2}{2a}\)
Substituting the given values, we get:
\(s = \frac{0^2 - 12^2}{2(-3)}\)
Now, let's calculate the distance:
\(s = \frac{-144}{-6}\)
\(s = 24\) meters
So, the motorcyclist will travel a distance of 24 meters while coming to rest.