A motorcycle racer falls off their motorbike while traveling at 102 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 4.6 m/s/s, how far (what displacement) would the rider travel before coming to a stop?
102 km/hr = 28.33 m/s
Now, as usual,
v = 28.33 - 4.6t
find t when v=0, and then
s = 28.33t - 2.3t^2
I'm still lost on what the answer is?
how did you get 28.33 m/s?
102 km/hr = 28.33 m/s from
102km/hr*1000m/km*1hr/3600sec= 28.33 m/s
If you are serious with this question, you need some tutoring on unit conversion very quickly.
Hi, im just curious and would appreciate it if someone replied. Would the answer be 87.24 m?
To find the distance (displacement) the rider travels before coming to a stop, we can use the following kinematic equation:
v^2 = u^2 + 2as
Where:
v is the final velocity (zero since the rider comes to a stop)
u is the initial velocity (102 km/hr converted to m/s)
a is the acceleration (negative since it decelerates)
s is the distance (displacement) we're trying to find
Let's break down the problem step by step.
1. Convert the initial velocity from km/hr to m/s:
Given: initial velocity, u = 102 km/hr
To convert km/hr to m/s, divide by 3.6:
u = 102 km/hr ÷ 3.6 = 28.33 m/s (rounded to two decimal places)
2. Use the kinematic equation to find the distance:
Given: final velocity, v = 0 m/s
Given: acceleration, a = -4.6 m/s^2
Given: initial velocity, u = 28.33 m/s
Let's substitute these values into the equation:
0^2 = (28.33)^2 + 2(-4.6)s
Simplifying the equation:
0 = 802.6689 - 9.2s
Rearrange the equation to solve for s:
9.2s = 802.6689
s = 802.6689 / 9.2
s ≈ 87.26 meters
Therefore, the rider would travel approximately 87.26 meters before coming to a stop.