1. Carlos jogs in a straight line at a constant speed of 1.5 m/s. He passes by Victoria, who, 10 seconds after Carlos had passed her, starts accelerating at a constant rate of 0.50 m/s^2
a) How much time after the passing of Carlos does it take Victoria to catch up to him?
b) What distance did they travel?
Assuming that Victoria was standing still when Carlos passed her, we have the following equations, calling t=0 when he passes her:
Carlos: s = 1.5t
Victoria: s = .25(t-10)^2
set them equal to find when they have traveled the same distance -- that is, when she passes him.
1.5t = .25(t^2 - 20t + 100)
1.5t = t^2/4 - 5t + 25
t^2/4 - 6.5t + 25 = 0
t^2 - 26t + 100 = 0
t = 13 ± √69
t = 4.7, 21.3
Since Victoria does not start till t=10, the only answer is 21.3, or 11.3 seconds after she started running.
Plug in times to get distances.
To tackle this problem, we need to understand the motion equations for Carlos and Victoria.
For Carlos, he is moving at a constant speed of 1.5 m/s. Since his speed is constant, we know that there is no acceleration. Hence, we can use the equation:
Distance = Speed × Time
For Victoria, she starts accelerating at a constant rate of 0.50 m/s², which means her velocity is changing over time. Since we know her acceleration, we can use the following equation to find her final velocity (v) after a certain time (t):
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Let's solve the problem step by step.
a) To find the time it takes for Victoria to catch up to Carlos, we need to consider that Carlos has already been jogging for 10 seconds when Victoria starts to accelerate.
Since Carlos has been moving at a constant speed of 1.5 m/s for 10 seconds, his distance traveled can be calculated as:
Carlos's Distance = Carlos's Speed × Carlos's Time
= 1.5 m/s × 10 s
= 15 meters
Now, let's calculate the time it takes for Victoria to catch up to Carlos.
We need to consider that Carlos has already covered a distance of 15 meters when Victoria starts.
To find the time it takes for Victoria to catch up to Carlos, we can use the equation:
Distance = Speed × Time
Let's assume the time it takes for Victoria to catch up is 't' seconds. Hence, her distance traveled would be:
Victoria's Distance = Victoria's Speed × t
At the time they both meet, Carlos's Distance = Victoria's Distance.
So, we can set up the following equation:
Carlos's Distance + Victoria's Distance = Victoria's Speed × t
15 meters + Victoria's Speed × t = Victoria's Speed × t
Since we are solving for time (t), we can rearrange the equation:
15 meters = Victoria's Speed × t
We know that Victoria starts with an initial speed of 0 m/s, and her acceleration is 0.50 m/s².
Using the equation v = u + at, we can solve for Victoria's final velocity (v).
Since her initial velocity (u) is 0, the equation simplifies to:
v = at
Given that her acceleration (a) is 0.50 m/s², we can substitute the values in to find her final velocity (v):
v = 0.50 m/s² × t
v = 0.50t
Now, we can substitute this value of v into our previous equation:
15 meters = Victoria's Speed × t
15 meters = 0.50t × t
15 meters = 0.50t²
We now have a quadratic equation. Simplifying further, we get:
0.50t² = 15 meters
Divide both sides by 0.50:
t² = 15 meters / 0.50
t² = 30
t = √30
t ≈ 5.50 seconds
Therefore, it takes approximately 5.50 seconds for Victoria to catch up to Carlos after Carlos passed her.
b) To find the distance they traveled, we can use either Carlos or Victoria's time, as they both travel the same distance when they meet.
Using Carlos's time of 10 seconds and his speed of 1.5 m/s, we can calculate his distance as:
Carlos's Distance = Carlos's Speed × Carlos's Time
= 1.5 m/s × 10 s
= 15 meters
Therefore, they traveled a distance of 15 meters.