Part A.) What is the final velocity, in meters per second, of a freight train that accelerates at a rate of 0.095 m/s2 for 8.75 min, starting with an initial velocity of 3.8 m/s?

a) A car starts from rest, speeds up with constant acceleration, and travels 720 meters in 9 seconds. What was the average velocity of the car in m/s? b) What was the final velocity of the car (in m/s)? c) Consider that the car started at rest. What was the change in velocity of the car (in m/s)? d) What was the acceleration of the car (in m/s2)?

You can just answer in a few sentences.

In class, we used a numerical method to determine the velocity of an object at 0s and = t = 2.00 s given that the object had started from rest at t accelerated at a rate given by a(t) = (8.00 m/s^)t². Here's what we did: 1. Divided t into four intervals of 0.5 s each. 2. For each interval, calculated the average a. These were the numerical values we obtained for each time interval: 1 0.25 0.50 i t (s) a; (m/s²) After summing the values according to: 2 0.75 4.50 3 1.25 12.50 lim ã‚At, 2 V₂v₁ = limast. t₁ At→0 we arrived at the approximate value v = 21.0 m/s. 4 1.75 24.50 a (m/s²) 32.00+ 24.00 16.00 8.00 0 ā₁ 0 ā2 az as 0.50 1.00 1.50 2.00 t (s) 21.33 m/s. We then proceeded to integrate to obtain the analytical solution, which was v = Using the same numerical method, divide t into eight and then sixteen intervals to calculate the approximate value for v at t = 2.00 s. This should demonstrate that with more sampling, the AUC (area under the curve) value converges toward the analytical…

5. DETERMINE THE STOPPING DISTANCES FOR A CAR WITH AN INITIAL SPEED OF 95km/h AND A HUMAN REACTION TIME OF 1.0 s, FOR AN ACCELERATION OF............. (A) a=-4.0 m/s², (B) a= -8.0 m/s² (Hint: First, convert 95 km/h to m/s)

A bike first accelerates from 0.0 m/s to 5.0 m/s in 4.5 s, then continues at this constant speed for another 4.5 s. What is the total distance traveled by the bike? Use sketches on velocity vs time and acceleration vs time graphs to make your case.

At a time t seconds, after it is thrown up in the air, a tomato is at a height (in meters) of f(t)= -4.9t^2+50t+3m A.) What is the average velocity of the tomato during the first 4 seconds? B.) Find (exactly) the instantaneous velocity of the tomato at t =4 C.) What is the acceleration at t =4?  D.) How high does the tomato go? E.)

In order to pass a physical education class at a university, a student must run 1.0 mi in 11.5 min. After running for 10.0 min (assume constant speed), she still has 660 yd to go. What constant acceleration does she need over the final 660 yd in order to make it? (Note: 1 yard is equal to 0.9144 meters.)  m/s2

a car travels on a straight, level road. a) Starting from rest, the car is going 42 ft/s (29 mi/h) at the end of 4.5 s. What is the car's average acceleration in ft/s2? b) In 3.5 more seconds, the car is going 84 ft/s (57 mi/h). What is the car's average acceleration for this time period? (c) The car then slows to 63 ft/s (43 mi/h) in 2.5 s. What is the average acceleration for this time period? (d) What is the overall average acceleration for the total time?

4. What was the initial velocity of an object that moved 100 m in 5.60 s reaching a final velocity of 15.0 m/s in that time? Was the object speeding up or slowing down? 5. A car traveling at 40 km/h accelerates at 2.3 m/s² for 2.7 s. How far has it traveled in that time? What is its final velocity?