During a track event two runners, Mary, and Alice, round the last turn and head into the final stretch with Mary a distance d = 2.0 m in front of Alice. They are both running with the same speed of v0= 5.0 m/s. When the finish line is a distance L= 45.0 m away from Alice, Alice accelerates at aA = 1.5 m/s2 until she catches up to Mary. Alice then continues at a constant speed until she reaches the finish line.
(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/s2)?
(e) What is Mary's velocity at the finish line (in m/s)?
5s
To find the answers to the given questions, we'll break down the problem into different stages and calculate the values step by step.
Let's start by finding the time it takes for Alice to catch up with Mary.
Step 1: Find the time it takes for Alice to catch up with Mary during her acceleration phase.
1. The initial distance between Mary and Alice is d = 2.0 m.
2. Alice's acceleration is aA = 1.5 m/s^2.
3. Alice is running at a constant speed of v0 = 5.0 m/s.
4. We need to find the time t1 it takes for Alice to catch up with Mary.
During the acceleration phase, Alice's position can be described using the equation:
dA = v0t1 + 0.5aAt1^2 (1)
where dA is the distance covered by Alice during the acceleration phase.
Since we know that Alice catches up with Mary, we can set dA = d:
d = v0t1 + 0.5aAt1^2 (2)
Rearranging equation (2), we get:
0.5aAt1^2 + v0t1 - d = 0 (3)
This is a quadratic equation in terms of t1. We can solve this equation using the quadratic formula:
t1 = (-v0 ± sqrt(v0^2 - 4(0.5aA)(-d))) / (2(0.5aA)) (4)
Substituting the given values into equation (4), we can find t1:
t1 = (-5.0 ± sqrt(5.0^2 - 4(0.5*1.5)(-2.0))) / (2(0.5*1.5))
Solving this equation will give us two possible values for t1. We choose the positive value, as time cannot be negative.
Step 2: Find the time it takes for Alice to reach the finish line after catching up with Mary.
1. The distance between Alice and the finish line is L = 45.0 m.
2. Alice's acceleration during this phase is 0 since she continues at a constant speed.
3. We need to find the time t2 it takes for Alice to reach the finish line after catching up with Mary.
To calculate t2, we can use the equation:
L = v0t2 (5)
Substituting the values into equation (5), we can find t2:
t2 = L / v0
Now, let's calculate the values:
(a) How long (in s) did it take Alice to catch up with Mary?
Substitute the values into equation (4) and solve for t1.
(b) How far (in m) did Alice still have to run when she just caught up to Mary?
Substitute the values of v0 and t1 into equation (1) to find dA. Subtract this value from L to get the remaining distance Alice has to run.
(c) How long (in s) did Alice take to reach the finish line after she just caught up to Mary?
Simply substitute the values of L and v0 into equation (5) to find t2.
(d) What is Mary's acceleration (in m/s^2)?
Since Mary accelerates all the way to the finish line and crosses at the same time as Alice, we can use equation (3) to find Mary's acceleration. Rearrange the equation to solve for aA.
(e) What is Mary's velocity at the finish line (in m/s)?
Substitute the values of v0, t1, and aA into equation (1) to find Mary's distance at the finish line.