A 109 kg fullback runs at the line of scrimmage.
(a) Find the constant force that must be exerted on him to bring him to rest in a distance of 1.1 m in a time interval of 0.22 s.
(b) How fast was he running initially?
someone answered this earlier but i do not understand how you got your answers since once needs to be in Newtons (part a) and the other in m/s (part b) Thanks!
All the equations you need are in my previous post
http://www.jiskha.com/display.cgi?id=1296751024
Use them and you will get the right units. I did part (b) first since it could be done in one step.
To find the answers, we can use the equations of motion. Let's assume that the acceleration of the fullback is constant.
(a) Finding the constant force:
We can use Newton's second law, which states that force is equal to mass multiplied by acceleration (F = m * a).
1. We are given:
Mass (m) = 109 kg
Distance (d) = 1.1 m
Time (t) = 0.22 s
2. To find acceleration:
We can use the equation of motion: d = (1/2) * a * t^2.
Rearranging the equation gives:
a = 2 * d / t^2
a = 2 * 1.1 m / (0.22 s)^2
a = 10 m/s^2
3. Now we can find the force required to bring the fullback to rest:
F = m * a
F = 109 kg * 10 m/s^2
F = 1090 N
So, the constant force required to bring the fullback to rest in a distance of 1.1 m in a time interval of 0.22 s is 1090 Newtons.
(b) Finding the initial speed:
We can use the equation of motion: v^2 = u^2 + 2 * a * d, where:
v = final velocity (0 m/s, as the fullback comes to rest)
u = initial velocity (what we need to find)
a = acceleration (from part a: 10 m/s^2)
d = distance (1.1 m)
1. Rearranging the equation gives:
u^2 = v^2 - 2 * a * d
u^2 = 0 - 2 * (10 m/s^2) * (1.1 m)
u^2 = -2 * 10 m^2/s^2 * 1.1 m
u^2 = -22 m^2/s^2
Note that the negative sign indicates that the initial velocity is in the opposite direction of the final velocity.
2. Taking the square root of both sides to solve for u:
u = √(-22 m^2/s^2)
u is an imaginary number, which is not physically meaningful in this context.
Therefore, it is not possible to determine the initial velocity of the fullback based on the given information.
To solve this problem, we can apply the concepts of Newton's second law and kinematics.
(a) Let's start with part (a), where we need to find the constant force exerted on the fullback to bring him to rest in a distance of 1.1 m in a time interval of 0.22 s.
We can use Newton's second law, which states that the force (F) exerted on an object is equal to the mass (m) of the object multiplied by its acceleration (a): F = m * a.
In this case, the object is the fullback, and we want to find the force, so we can rearrange the equation to solve for F: F = m * a.
We can find the acceleration using kinematics. The final velocity of the fullback is 0 m/s since he comes to rest. The initial velocity is unknown. The distance traveled is 1.1 m, and the time interval is 0.22 s.
The equation we can use to find acceleration is: v_f = v_i + a * t, where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time interval.
Substituting the known values: 0 = v_i + a * 0.22.
Since the fullback comes to rest, the final velocity (v_f) is 0. So the equation becomes: 0 = v_i + 0.22a.
We can rearrange this equation to solve for a: a = -v_i / 0.22.
Now, we can substitute this value of acceleration (a) and mass (m) into the force equation: F = m * a.
F = 109 kg * (-v_i / 0.22).
Simplifying, we get: F = -500v_i N.
So, the constant force that must be exerted on the fullback to bring him to rest is -500v_i Newtons.
(b) Moving on to part (b), we need to find the initial velocity of the fullback.
We can use the same kinematic equation: v_f = v_i + a * t.
Since the final velocity (v_f) is 0 (the fullback comes to rest), the equation becomes: 0 = v_i + (-v_i / 0.22) * 0.22.
Simplifying, we get: 0 = v_i - v_i.
This implies that the initial velocity (v_i) is equal to 0.
Therefore, the fullback was initially running with a velocity of 0 m/s.