In an historical movie, two knights on horseback start from rest 79.0 m apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.319 m/s2, while Sir Alfred's has a magnitude of 0.391 m/s2. Relative to Sir George's starting point, where do the knights collide?
Dear student,
write the equation of motion for each horse; in general you have
x(t)=x0+v0 t + 1/2 at^2
x(t)=position as a function of time
x0=initial position
v0=initial speed (which here is zero for both horses)
a=acceleration
t=time
for George, we have:
x(t)=0+0+1/2 (0.22)t^2
for Arthur, we have:
x(t)=70+0-1/2(0.3)t^2 (accel is negative since it is in the opposite direction of George's
they meet when the have the same value of x:
0.11t6^2=70-0.15t^2
0.26t^2=70 or t=16.4 s
plug this into the george equation to get:
x(t=16.4s)=0.11(16.4)^2=29.6 m
Eevee,
To find the distance from Sir George's starting point where the knights collide, we need to determine the time it takes for them to meet.
We can start by using the formula to calculate the time it takes for an object to reach a certain distance with constant acceleration:
s = ut + (1/2)at^2
Where:
s is the distance,
u is the initial velocity (which is zero in this case),
a is the acceleration, and
t is the time.
We can rearrange the formula to solve for time:
t = sqrt(2s/a)
For Sir George's acceleration:
a₁ = 0.319 m/s²
s₁ = 79.0 m
t₁ = sqrt(2s₁/a₁)
= sqrt(2 * 79.0 / 0.319)
≈ sqrt(498.4289)
≈ 22.33 s (rounded to two decimal places)
Now, let's calculate the time it takes for Sir Alfred to reach the same distance from Sir George's starting point:
For Sir Alfred's acceleration:
a₂ = 0.391 m/s²
t₂ = sqrt(2s₁/a₂)
= sqrt(2 * 79.0 / 0.391)
≈ sqrt(405.876
Since Sir George and Sir Alfred ride toward each other, they will meet at the same time, which is approximately 22.33 seconds (rounded).
To determine the distance from Sir George's starting point where the knights collide, we can use the formula:
distance = s + ut + (1/2)at^2
Since Sir George's initial velocity is zero, the formula simplifies to:
distance = (1/2)at^2
Using t = 22.33 s and a = 0.319 m/s²:
distance = (1/2)(0.319)(22.33)^2
≈ 80.17 m (rounded to two decimal places)
Therefore, the knights will collide approximately 80.17 meters from Sir George's starting point.
To determine where the knights collide, we need to calculate the time it takes for them to meet. Since we know their initial positions, accelerations, and that they start from rest, we can use the kinematic equation:
\(d = v_0t + \frac{1}{2}at^2\)
First, let's calculate the time it takes for Sir George to meet Sir Alfred. We'll refer to Sir George's position as positive and Sir Alfred's as negative:
For Sir George:
\(d_{\text{George}} = 0\) (starting from rest)
\(a_{\text{George}} = 0.319 \, \text{m/s}^2\)
For Sir Alfred:
\(d_{\text{Alfred}} = -79.0 \, \text{m}\) (starting from rest and moving in the opposite direction)
\(a_{\text{Alfred}} = -0.391 \, \text{m/s}^2\)
Using the kinematic equation, we have:
\(0 = 0 \cdot t + \frac{1}{2} \cdot 0.319 \cdot t^2\)
Simplifying the equation, we get:
\(0.159 \cdot t^2 = 0\)
Since the product of the non-zero acceleration and the time squared is zero, we can conclude that \(t = 0\).
This means that Sir George and Sir Alfred meet at the same instant they start moving. Now, to find their position at the moment of collision, we substitute this time into the kinematic equation for either knight.
Let's use the equation for Sir Alfred:
\(d_{\text{Alfred}} = -79.0 + 0 \cdot t + \frac{1}{2} \cdot (-0.391) \cdot t^2\)
Since \(t = 0\), the equation simplifies to:
\(d_{\text{Alfred}} = -79.0 \, \text{m}\)
Therefore, relative to Sir George's starting point, the knights collide at a distance of -79.0 meters. The negative sign indicates that the collision occurs in the opposite direction from Sir George.