A motorcycle racer falls off their motorbike while travelling at 143 km/hr and skids in a straight line down the track. If the friction with the track decelerates the rider at a uniform rate of 3.8 m/s/s, how long (what time) would it take for the rider to come to a stop?
These last two are just like the one I answered. Come back with your work if you get stuck.
can you show the formula on how to figure this out?
change initial velocity to m/s
vf=vi+vi*t+1/2 a t^2
vf=0, vi=initial velocity
a=-3.8, so you solve for time t. It is a quadratic equation, I recommend put it in this form...
1/2 at^2+Vi*t-Vi=0 and solve
To find the time it takes for the rider to come to a stop, we can use the equation of motion that relates acceleration, initial velocity, final velocity, and time:
v = u + at
Where:
v = final velocity (0 m/s, since the rider comes to a stop)
u = initial velocity of the rider (143 km/hr = 39.72 m/s)
a = acceleration (deceleration in this case, -3.8 m/s²)
t = time
Rearranging the equation to solve for time, we have:
t = (v - u) / a
Substituting the given values, we get:
t = (0 - 39.72) / (-3.8)
Calculating this equation gives us:
t = 10.45 seconds
Therefore, it would take the rider approximately 10.45 seconds to come to a stop after falling off the motorbike.