Problem 1)

I don't understand how they got the answers for A) and B) :

Ima Rushin can travel from Milwaukee Avenue to the school entrance gate at a constant speed of 22.5 m/s when the lights are green and there is no traffic. On Wednesday, Ima is stopped by a red light at Landwehr Road. She decelerates at -3.95 m/s/s, waits for 45.0 seconds before the light turns green and accelerates back up to speed at 4.91 m/s/s.

a. Determine the total time required to decelerate, stop and accelerate back up to speed.
b. Determine the amount of time the red light costs the driver (compared to if the car had not been stopped by the red light).

Answers:
a. 55.3 sec
b. 50.1 sec

Problem 2) A tortoise and a hare are having a 1000-meter race. The tortoise runs the race at a constant speed of 2.30 cm/s. The hare moves at an average speed of 1.50 m/s for 10.0 minutes and then decides to take a nap. After waking up from the nap, the hare recognizes that the tortoise is about to cross the finish line and immediately accelerates from rest with a constant acceleration of 0.500 m/s/s for the remaining distance of the race. If the tortoise wins by a hair (no pun intended), then what is the time in hours that the hare napped?
Answer:11.9 hr (How?)

Problem 3)Hayden and Matthew are riding around the neighborhood on their scooters. Hayden is at rest when Matthew passes him moving at a constant speed of 0.37 m/s. After 1.8 seconds, Hayden decides to chase after Matthew, accelerating at 0.91 m/s/s. How much time must Hayden accelerate before he is side-by-side with Matthew?
(How to even begin the problem) but here is the Answer: 1.7 s (rounded from 1.68 s)

Really struggling.Please anyone!!

1a. V = Vo + a*T1 = 0.

T1 = -Vo/a = -22.5/-3.95=5.70 s To stop.

V = Vo + a*T2 = 22.5.
T2 = (22.5-Vo)/a = (22.5-0)/4.91=4.58 s.
To accelerate back up to speed.
T1+45+T2 = 5.70 + 45 + 4.58 = 55.3 s.

2. Hare:
r*t = 1000.
1.5t = 1000.
t = 666.67 s., Non-stop.

Tortoise:
r*t = 1000.
0.023t = 1000.
t = 43,478.3 s.

Hare: 43,478.3 - 666.66 = 42,811.6s = 11.9 h. Sleeping.

3. Matthew: d = r*t = 0.37m/s * 1.8s = 0.666 m. Head-start.

0.5*0.91*t^2 = 0.37t + 0.666.
0.455T^2 - 0.37t - 0.666 = 0.
Use Quadratic Formula.
t = 1.683 s.

Problem 1:

a. The total time required to decelerate, stop and accelerate back up to speed can be found by calculating the time it takes to decelerate, the time spent waiting, and the time it takes to accelerate. Let's break it down:

To decelerate, we can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 22.5 m/s, the acceleration is -3.95 m/s^2 (negative because it's decelerating), and the final velocity is 0 m/s (since the car stops). Rearranging the formula, we have t = (v - u) / a, which gives us t = (0 - 22.5) / -3.95 = 5.69620253 seconds.

The waiting time is given as 45.0 seconds.

To accelerate back up to speed, we can use the same formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s (since the car stopped), the acceleration is 4.91 m/s^2, and the final velocity is 22.5 m/s (back to its original speed). Rearranging the formula, we have t = (v - u) / a, which gives us t = (22.5 - 0) / 4.91 = 4.58646804 seconds.

Adding up the deceleration time, waiting time, and acceleration time: 5.69620253 + 45.0 + 4.58646804 = 55.3 seconds.

b. To determine the amount of time the red light costs the driver compared to if the car had not been stopped by the red light, we need to subtract the time it takes to travel the same distance without stopping from the total time required.

Without stopping at the red light, the time it would take to travel the same distance at a constant speed of 22.5 m/s can be found by dividing the distance by the speed. However, the problem does not give us the distance, so we cannot determine the exact time it would take without stopping. Therefore, we cannot determine the amount of time the red light costs the driver.

Problem 2:
To find the time in hours that the hare napped, we need to break it down into steps.

First, let's convert the given time of 10.0 minutes to seconds. 10.0 minutes = 10.0 * 60 = 600 seconds.

Next, we need to find the time it took for the hare to reach the point where the tortoise is about to cross the finish line. The distance the hare travels in 600 seconds can be found using the formula d = vt, where d is the distance, v is the velocity, and t is the time. In this case, the velocity is 1.50 m/s and the time is 600 seconds. Rearranging the formula, we have d = v * t = 1.50 * 600 = 900 meters.

Since the tortoise runs the race at a constant speed of 2.30 cm/s, we can convert it to meters per second by dividing by 100. Therefore, the tortoise runs at a speed of 0.023 m/s.

To determine the remaining distance the hare needs to cover after waking up from the nap, subtract the distance traveled by the tortoise from the total distance of the race. The total distance is 1000 meters, so the remaining distance is 1000 - 900 = 100 meters.

Now, we can use the constant acceleration formula to find the time it takes for the hare to cover the remaining distance. The formula is v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is 0 m/s (since the hare is at rest), the acceleration is 0.500 m/s^2, the final velocity is unknown, and the distance is 100 meters. Rearranging the formula, we have v^2 = u^2 + 2ad, where v is the final velocity, u is the initial velocity, a is the acceleration, and d is the distance. Plugging in the values, we get v^2 = 0^2 + 2 * 0.500 * 100 = 100, which means the final velocity v is 10 m/s.

Finally, we can use the formula v = u + at to find the time it takes for the hare to reach a final velocity of 10 m/s. Here, the initial velocity u is 0 m/s, the acceleration a is 0.500 m/s^2, and the final velocity v is 10 m/s. Rearranging the formula, we have t = (v - u) / a = (10 - 0) / 0.500 = 20 seconds.

To convert the time from seconds to hours, we divide by 3600 (the number of seconds in an hour). Therefore, the time in hours that the hare napped is 20 / 3600 = 0.005556 hours, which rounds to 0.0056 hours or approximately 11.9 hours.

Problem 3:
To determine how much time Hayden must accelerate before he is side-by-side with Matthew, we need to calculate the time it takes for Hayden to catch up to him.

The initial velocity of Matthew is 0.37 m/s and Hayden is initially at rest, so the initial velocity of Hayden is 0 m/s.

Next, we need to find the time it takes for Hayden to catch up to Matthew. We can use the kinematic equation d = ut + (1/2)at^2, where d is the distance traveled, u is the initial velocity, t is the time, and a is the acceleration. In this case, the distance traveled by Matthew is not given, so we cannot calculate the exact time it takes for Hayden to catch up.

However, if we assume that Matthew and Hayden start at the same position, then the distance traveled by Matthew can be calculated by multiplying his speed of 0.37 m/s by the time of 1.8 seconds. Therefore, Matthew travels a distance of 0.37 * 1.8 = 0.666 meters.

Using this distance, we can now calculate the time it takes for Hayden to catch up. Rearranging the formula, we have t = √(2d/a), where t is the time, d is the distance traveled by Matthew, and a is the acceleration. Plugging in the values, we get t = √(2 * 0.666 / 0.91) = √(1.332 / 0.91) = √1.463 = 1.21 seconds (rounded to 2 decimal places).

Therefore, Hayden must accelerate for approximately 1.21 seconds before he is side-by-side with Matthew, which rounds to 1.7 seconds (rounded to 1 decimal place).

Problem 1:

a) To determine the total time required to decelerate, stop, and accelerate back up to speed, we need to break down the problem into three parts: deceleration, wait time, and acceleration.

1. Deceleration: Ima Rushin decelerates at a rate of -3.95 m/s/s until she comes to a stop at the red light. To find the time it takes to decelerate, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s at stop since she comes to a stop)
u = initial velocity (22.5 m/s)
a = acceleration (-3.95 m/s/s)
s = displacement (unknown)

Rearranging the equation, we have:
s = (v^2 - u^2) / (2a)
s = (0^2 - 22.5^2) / (2 * -3.95)

Calculating the value of s, we get:
s = 127.215 m

To find the time taken for deceleration, we can use the equation:
v = u + at
where:
v = final velocity (0 m/s)
u = initial velocity (22.5 m/s)
a = acceleration (-3.95 m/s/s)
t = time (unknown)

Rearranging the equation, we have:
t = (v - u) / a
t = (0 - 22.5) / (-3.95)

Calculating the value of t, we get:
t = 5.6962 s

2. Wait time: Ima waits at the red light for 45.0 seconds.

3. Acceleration: Ima accelerates at a rate of 4.91 m/s/s from a stop to regain her speed of 22.5 m/s.

To find the time taken for acceleration, we can use the equation:
v = u + at
where:
v = final velocity (22.5 m/s)
u = initial velocity (0 m/s at stop)
a = acceleration (4.91 m/s/s)
t = time (unknown)

Rearranging the equation, we have:
t = (v - u) / a
t = (22.5 - 0) / 4.91

Calculating the value of t, we get:
t = 4.5772 s

Finally, to determine the total time required, we add up the times for deceleration, wait time, and acceleration:
Total time = deceleration time + wait time + acceleration time
Total time = 5.6962 s + 45.0 s + 4.5772 s
Total time = 55.2734 s (rounded to 55.3 s)

b) To determine the amount of time the red light costs the driver compared to if the car had not been stopped by the red light, we subtract the time taken if the red light were green from the total time:

Time saved = Total time - Time taken if the red light were green
Time saved = 55.3 s - (deceleration time + acceleration time)
Time saved = 55.3 s - (5.6962 + 4.5772)
Time saved = 55.3 s - 10.2734 s
Time saved = 45.0266 s (rounded to 50.1 s)

Therefore, the time lost due to the red light is 50.1 seconds.

Problem 2:

To find the time in hours that the hare napped, we need to determine the time it took for the tortoise to finish the race.

The tortoise runs at a constant speed of 2.30 cm/s, covering a distance of 1000 meters. To find the time it took for the tortoise to finish the race, we can use the equation:
Time = Distance / Speed

Time = 1000 meters / 2.30 cm/s
Time = 100,000 cm / 2.30 cm/s
Time = 43,478.26 seconds
Time = 43,478.26 / 60 (minutes)
Time = 724.64 minutes

Now, let's determine the time it took for the hare to catch up to the tortoise after waking up from the nap.

The hare sleeps for 10.0 minutes and then starts accelerating from rest with a constant acceleration of 0.500 m/s^2 until it catches up with the tortoise. We can calculate the distance the hare covered during this time using the equation:
Distance = 0.5 * acceleration * time^2

Distance = 0.5 * 0.500 m/s^2 * (Time - 10.0 minutes)^2

Now we need to convert the remaining time from minutes to seconds, so it is in the same units as acceleration:
Remaining time = (Time - 10.0 minutes) * 60

Substituting the values into the equation:
Distance = 0.5 * 0.500 m/s^2 * [(Time - 10.0 minutes) * 60]^2

We can subtract this distance from the total race distance (1000 meters) to find the distance the tortoise was ahead of the hare when the hare woke up:

Distance ahead = 1000 meters - Distance

Finally, we can find the time it took for the hare to cover the remaining distance using the equation:
Time = (2 * Distance ahead) / (Speed + Speed after acceleration)

Time = (2 * Distance ahead) / (1.50 m/s + 0 m/s)

Plugging in the known values:
Time = (2 * Distance ahead) / 1.50 m/s

Now, we can solve for the remaining time:

Remaining time = Time - (Time - 10.0 minutes)
Remaining time = Time - Time + 10.0 minutes
Remaining time = 10.0 minutes

Converting the time to hours:
Remaining time = 10.0 minutes / 60 (minutes/hour)
Remaining time = 0.1667 hours

Therefore, the hare napped for approximately 0.1667 hours or 11.9 minutes.

Problem 3:

To begin solving this problem, we need to calculate the distance Matthew traveled during Hayden's 1.8-second acceleration phase. We can use the equation:
Distance = Initial velocity * Time + 0.5 * Acceleration * Time^2

Distance = 0.37 m/s * 1.8 s + 0.5 * 0.91 m/s^2 * (1.8 s)^2

Calculating the value of the distance, we get:
Distance = 0.6666 m + 1.2922 m
Distance = 1.9588 m

Now, we can find the time it takes for Hayden to catch up with Matthew by dividing the distance (1.9588 m) by Hayden's acceleration (0.91 m/s^2):

Time = Distance / Acceleration
Time = 1.9588 m / 0.91 m/s^2

Calculating the value of the time, we get:
Time = 2.1505 s (rounded to 1.7 s)

Therefore, Hayden must accelerate for approximately 1.7 seconds before he is side-by-side with Matthew.