9)You are taking a road trip to the beach for Spring Break. You leave at 12:00pm. At 12:10 you stop for lunch 10 miles down the road. You get back on the road and travel 75 more miles and realize that you need to stop for gas. You notice that it is now 1:10pm. At this point in the trip, what is your average velocity in miles per hour?
10)An airplane must achieve a velocity of 71 m/s to takeoff. If the runway is 1000m long, at what speed must the plane constantly accelerate?
0.071 m/s^2
Checking your answer for 10:
For an initial velocity of 0, final velocity can be expressed as:
V = SQRT(2 * acceleration * distance)
Using your answer of 0.071 m/s:
V = SQRT(2 * (0.071 m/s^2) * 1000m)
V = SQRT( 142.0 m^2/s^2)
V = 11.92 m/s
Uh-oh! Not enough velocity by the end of the runway!
2.52 for #10
Yes!
2.52 m/s^2 is what I got also.
To find the average velocity, you need to calculate the total distance traveled and the total time taken.
For question 9:
1. Distance traveled = (10 miles + 75 miles) = 85 miles
2. Time taken = 1:10pm - 12:00pm = 1 hour and 10 minutes = 1.17 hours
Therefore, the average velocity is:
Average velocity = Distance / Time = 85 miles / 1.17 hours = 72.65 miles per hour
For question 10:
For the plane to achieve a velocity of 71 m/s with a runway length of 1000m, we need to calculate the constant acceleration required.
1. The equation to calculate final velocity with constant acceleration is: v = u + at
where v = final velocity, u = initial velocity, a = acceleration, t = time.
In this case, final velocity (v) is 71 m/s, initial velocity (u) is 0 m/s, and the time (t) is not given.
2. The equation to calculate displacement (s) with constant acceleration is: s = ut + (1/2)at^2
where s = displacement.
3. Since the runway length is the displacement, we can equate it to the above equation:
1000m = (1/2)(0 + a)(t^2)
Simplifying: 2000 = a(t^2)
4. We also know that the final velocity (71 m/s) is related to the acceleration and time:
71 = 0 + a(t)
Solving the above two equations simultaneously, we can find the value of t.
5. Rearranging the second equation, we get: t = 71/a
6. Substituting t in the first equation, we get: 2000 = a((71/a)^2)
Simplifying: 2000 = 71^2
Taking the square root of both sides: a = sqrt(2000) ≈ 44.72 m/s^2
Therefore, the plane must constantly accelerate at approximately 44.72 m/s^2 to achieve a velocity of 71 m/s on a 1000m long runway.