The motorcycle and rider are sliding with the same acceleration as found in Problem 4. If the motor- cycle and rider have been sliding for 4.55 s, what will be the force applied to the motorcycle and the rider when they strike the side barrier and come to rest in another 0.530 s?
To find the force applied to the motorcycle and rider when they strike the side barrier and come to rest, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
First, we need to calculate the acceleration of the motorcycle and rider. Since their acceleration is given in Problem 4, we will use the previous value.
Next, we need to find the mass of the motorcycle and rider. Let's assume the combined mass of the motorcycle and rider is represented by m. We will also assume that the mass is distributed evenly between the motorcycle and rider.
To find the force, we can use the equation:
F = m * a
Now we can calculate the force.
1. Calculate the acceleration:
Use the value given in Problem 4.
2. Calculate the mass:
Let's assume the combined mass of the motorcycle and rider is 200 kg. Since the mass is distributed evenly, the mass of the motorcycle would be 100 kg and the mass of the rider would also be 100 kg.
3. Calculate the force:
F = m * a
F = (100 kg + 100 kg) * acceleration
4. Calculate the time taken to come to rest after hitting the barrier:
The total time is given as 4.55 s + 0.530 s.
5. Calculate the deceleration:
Deceleration = Final Velocity / Time
6. Calculate the force upon coming to rest:
Force = m * deceleration
Please provide the acceleration from the previous problem (Problem 4), and I will continue with the calculations.
To find the force applied to the motorcycle and the rider when they strike the side barrier, we need to use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Given that the motorcycle and rider have been sliding with the same acceleration, we need to find the acceleration first. We can do this by using the equation:
\[a = \dfrac{\Delta v}{\Delta t}\]
Where:
a = acceleration
\(\Delta v\) = change in velocity
\(\Delta t\) = time interval
The problem doesn't mention any change in velocity, so we can assume that the initial velocity is zero (since they come to rest at the end). Therefore, the change in velocity is just the final velocity.
Next, we can use the equation of motion:
\[v = u + at\]
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Since the motorcycle and rider come to rest, the final velocity (v) is zero. We can rearrange the equation to solve for acceleration:
\[a = \dfrac{v - u}{t}\]
Substituting the given values, we have:
\[a = \dfrac{0 - 0}{0.530 s}\]
Since the numerator is zero, the acceleration is zero. This means that there is no force being applied to the motorcycle and rider during the time they come to rest.
Therefore, the force applied to the motorcycle and the rider when they strike the side barrier and come to rest is zero.