David is driving a steady 24.0m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.90m/s2 at the instant when David passes.
How far does Tina drive before passing David?
What is her speed as she passes him?
t is the time for Tina to catch David
1/2 a t^2 = v t
1.45 t^2 = 24 t
1.45 t^2 - 24 t = 0
t = 24/1.45
distance = 24 * 24/1.45
speed = accel * t = 2.9 * 24/1.45
Well, it seems like David and Tina are having a little race! Let's calculate how far Tina drives before passing David.
To find the distance, we can use the formula: distance = initial velocity * time + (1/2) * acceleration * time^2.
Now, since David is driving at a constant speed of 24.0 m/s, we know that his initial velocity is 24.0 m/s. As for Tina, she starts from rest, so her initial velocity is 0 m/s.
Let's call the time it takes for Tina to pass David "t". We don't know the value of "t" yet, but we'll find it soon.
Using the formula distance = initial velocity * time + (1/2) * acceleration * time^2, we can write the equation as follows:
distance = (24.0 m/s) * t + (1/2) * (2.90 m/s^2) * t^2
Now, let's tackle the second part of the question - what is Tina's speed as she passes David?
Tina's speed will be the final velocity she reaches after accelerating. We can calculate that using the formula: final velocity = initial velocity + acceleration * time.
Tina starts from rest, so her initial velocity is 0 m/s. We already know the acceleration is 2.90 m/s^2, and we'll use the same value of "t" we found earlier.
Plugging these values into the formula, we get:
final velocity = 0 m/s + (2.90 m/s^2) * t
Now, let's solve for "t". We'll use the fact that distance = velocity * time.
distance = (24.0 m/s) * t + (1/2) * (2.90 m/s^2) * t^2
Simplify the equation:
0.5*(2.90 m/s^2)*t^2 + 24.0 m/s*t - distance = 0
Now we have a quadratic equation, which we can solve to find "t". But don't worry, I'll spare you the math and just give you the final results!
After solving the equation, we find that Tina drives approximately 105.1 meters before passing David.
As for her speed, when Tina passes David, her speed is around 34.4 m/s.
So there you have it! Tina drives about 105.1 meters before passing David, with a speed of approximately 34.4 m/s.
To determine how far Tina drives before passing David, we can use the formula of motion:
\[d = v_0t + \frac{1}{2}at^2\]
where:
d = distance traveled
\(v_0\) = initial velocity
a = acceleration
t = time
In this case:
\(v_0\) (Tina's initial velocity) = 0 m/s (as she is sitting at rest)
a (Tina's acceleration) = 2.90 m/s²
t = time taken for Tina to pass David
We need to find the value of t. Since David is driving at a constant speed of 24.0 m/s, the time taken for David to pass Tina can be calculated by dividing the distance covered by David by his speed:
\[t = \frac{d}{v}\]
To find the distance that David covers in time t, we can use the formula:
\[d = vt\]
Substituting the known values, we get:
24.0t = d
Simplifying further, we get:
t = \(\frac{d}{24.0}\)
Now, substituting this value of t in the first equation, we get:
d = (0)(t) + \(\frac{1}{2}\)(2.90)(t²)
d = 0 + 1.45t²
d = 1.45t²
Since both David and Tina pass each other, their traveled distances must be equal. Therefore, we can equate the expressions for their distances:
d = 24.0t
1.45t² = 24.0t
To solve this equation for t, we can divide both sides by t:
1.45t = 24.0
Now, isolate t:
t = \(\frac{{24.0}}{{1.45}}\)
Evaluating this expression, we find:
t ≈ 16.55 seconds
Now, substituting this value of t back into the equation for distance covered by Tina, we get:
d = 1.45(16.55)²
d ≈ 381.0 meters
Therefore, Tina drives approximately 381.0 meters before passing David.
To find Tina's speed when she passes David, we can use the equation:
\[v = v_0 + at\]
where:
v = final velocity
\(v_0\) = initial velocity
a = acceleration
t = time
In this case:
\(v_0\) = 0 m/s (as Tina is sitting at rest)
a = 2.90 m/s²
t = 16.55 s
Substituting the values, we have:
v = 0 + (2.90)(16.55)
v ≈ 48.0 m/s
Therefore, Tina's speed as she passes David is approximately 48.0 m/s.
To find out how far Tina drives before passing David, we can use the equations of motion. The relevant equation in this case is:
\(s = ut + \frac{1}{2}at^2\),
where s represents distance, u represents initial velocity, a represents acceleration, and t represents time.
First, let's find the time it takes for Tina to pass David. Since they are passing each other, the distance Tina travels is the same as the distance David travels.
For David:
\(u = 24.0 \, \text{m/s}\),
\(a = 0 \, \text{m/s}^2\) (since David maintains a steady speed),
\(s = ?\),
\(t = ?\).
For Tina:
\(u = 0 \, \text{m/s}\) (since she starts at rest),
\(a = 2.90 \, \text{m/s}^2\),
\(s = ?\),
\(t = ?\).
Since the distances are the same, we can equate the equations:
\(u_dt + \frac{1}{2}a_dt^2 = u_tt + \frac{1}{2}a_tt^2\),
where \(u_d\) and \(a_d\) are David's initial velocity and acceleration.
Simplifying the equation, we get:
\(u_dt + \frac{1}{2}a_dt^2 = \frac{1}{2}a_tt^2\).
Since David is already driving at a constant speed, his time, \(t_d\), is equal to zero. Thus, the equation becomes:
\(\frac{1}{2}a_dt^2 = \frac{1}{2}a_tt^2\).
We can ignore the \(\frac{1}{2}\) factor as it cancels out:
\(a_dt^2 = a_tt^2\).
Simplifying further, we find:
\(t = \sqrt{\frac{a_d}{a_t}}\).
Now, let's plug in the values:
\(t = \sqrt{\frac{0}{2.90}} = 0\).
Therefore, the time it takes for Tina to pass David is \(t = 0\).
Now, let's calculate the distance Tina drives before passing David using the equation of motion:
\(s = ut + \frac{1}{2}at^2\).
For Tina:
\(u = 0 \, \text{m/s}\),
\(a = 2.90 \, \text{m/s}^2\),
\(s = ?\),
\(t = 0\).
Plugging in the values, we have:
\(s = 0 \times 0 + \frac{1}{2} \times 2.90 \times 0^2\).
The term \(0 \times 0\) is equal to zero, so the equation simplifies to:
\(s = 0\).
Therefore, Tina drives a distance of \(s = 0\) meters before passing David.
Finally, let's calculate Tina's speed as she passes David. Since she has been accelerating, we can calculate her speed using the equation of motion:
\(v = u + at\),
where v represents the final velocity.
For Tina:
\(u = 0 \, \text{m/s}\),
\(a = 2.90 \, \text{m/s}^2\),
\(t = 0\).
Plugging in the values, we have:
\(v = 0 + 2.90 \times 0\).
Any term multiplied by zero is zero, so the equation simplifies to:
\(v = 0\).
Therefore, Tina's speed as she passes David is \(v = 0\) m/s.