During a track event two runners, Mary, and Alice, round the last turn and head into the final stretch with Mary a distance d = 4.0 m in front of Alice. They are both running with the same speed of v0= 7.0 m/s. When the finish line is a distance L= 45.0 m away from Alice, Alice accelerates at aA = 0.5 m/s2 until she catches up to Mary. Alice then continues at a constant speed until she reaches the finish line.
physics - tini, Wednesday, September 11, 2013 at 11:25am
(a) How long (in s) did it take Alice to catch up with Mary?
(b) How far (in m) did Alice still have to run when she just caught up to Mary?
(c) How long (in s) did Alice take to reach the finish line after she just caught up to Mary?
Mary starts to accelerate when Alice just catches up to her, and accelerates all the way to the finish line and crosses the line exactly when Alice does. Assume Mary's acceleration is constant.
(d) What is Mary's acceleration (in m/
(e) What is Mary's velocity at the finish line (in m/s)?
v₀=7m/s; d=4 m; L 45 m; a= 0.5 m/s².
(v₀t + at²/2) - (v₀t) = d
t =sqrt (2d/a) = sqrt(2•4/0.5) = 2.83 s
(a) t=2.83 s
v₁=v₀+at = 7 + 0.5•2.83 = 8.42 m/s
x = v₀t+ at²/2 = 7•2.83 + 0.5•2.83²= 21.8 m
(b) s = L – x = 42 – 21.8 = 23.2 m
(c)
t₁ = s/v₁ = 23.2 /8.42 = 2.76 s.
(d)
s=v₀t₁+ a₁t₁²/2
a₁ =2{s- v₀t₁)/t₁² =
=2(23.2 - 7•2.76)/2.76² =1/02 m/s²
(e)
v₂=v₀+a₁t₁ =7+1.02•2.76 = 9.81 m/s
To solve this problem, we can break it down into a few steps:
Step 1: Determine the time it takes for Alice to catch up with Mary.
To find the time it takes for Alice to catch up with Mary, we need to consider their initial positions and velocities. Mary is initially 4.0 m ahead of Alice, and they both have the same speed of 7.0 m/s.
Let's define the time it takes for Alice to catch up with Mary as t_catch. During this time, Alice's initial position will be 4.0 m behind Mary, and she will be continuously accelerating until she catches up.
Using the equations of motion, we can write the position of Alice as:
x_Alice = 4.0 m + (1/2) * a_Alice * t_catch^2
where a_Alice represents Alice's acceleration.
Since Alice accelerates at a rate of 0.5 m/s^2, we can substitute this value into the equation:
x_Alice = 4.0 m + (1/2) * 0.5 m/s^2 * t_catch^2
x_Alice = 4.0 m + 0.25 m/s^2 * t_catch^2
Now let's consider Mary's position. Mary will continue running at a constant speed until she is caught by Alice. The position of Mary at time t_catch is:
x_Mary = v_Mary * t_catch
where v_Mary represents Mary's velocity. In this case, since Mary's acceleration is not given, we assume it to be zero. Therefore, Mary runs at a constant speed of 7.0 m/s.
Now we can set up an equation equating the positions of Alice and Mary at time t_catch:
x_Alice = x_Mary
4.0 m + 0.25 m/s^2 * t_catch^2 = 7.0 m/s * t_catch
To solve this equation for t_catch, we can rearrange it to:
0.25 m/s^2 * t_catch^2 - 7.0 m/s * t_catch + 4.0 m = 0
This is a quadratic equation in terms of t_catch. We can solve it using the quadratic formula.
Step 2: Determine the distance Alice had left to run when she caught up with Mary.
Once we find t_catch, we can calculate the distance Alice had left to run when she caught up with Mary. To do this, we need to find Alice's position at time t_catch.
We can use the same equation we used before to find Alice's position:
x_Alice = 4.0 m + 0.25 m/s^2 * t_catch^2
Step 3: Determine the time it took Alice to reach the finish line after catching up with Mary.
Since Mary's acceleration is not given, we assume it to be constant. This means Mary's speed will increase from 7.0 m/s to some final velocity v_final. Mary accelerates from the moment Alice catches up with her until they both cross the finish line together. We need to find the time it takes for Mary to reach the finish line after being caught up by Alice.
To find this time, we can use the equation of motion for uniform acceleration:
x = x_0 + v_0 * t + (1/2) * a * t^2
where x is the distance, x_0 is the initial position, v_0 is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial position x_0 is the distance Alice had left to run when she caught up with Mary. The initial velocity v_0 is the velocity at which Alice catches up with Mary. The final position x is the distance to the finish line, L. We know the distance to the finish line is 45.0 m.
So we can write the equation as:
L = x_Alice + v_Alice * t_finish + (1/2) * a_Mary * t_finish^2
where v_Alice is Alice's velocity at the time when she caught up with Mary, and a_Mary is Mary's acceleration.
Step 4: Determine Mary's acceleration and velocity at the finish line.
Since we know the distance to the finish line and the time it takes for Mary to reach the finish line after being caught up by Alice, we can solve the equation from Step 3 to find Mary's acceleration. Once we have Mary's acceleration, we can calculate her velocity at the finish line.
I hope this step-by-step explanation helps you understand how to approach and solve this problem. If you have any further questions, please let me know!