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- The acceleration of a bus is given by a, (t) = at, where a = 1.25 m/s³ is a constant. ▾ Part A If the bus's velocity at time ₁1.15 s is 4.90 m/s, what is its velocity at time t₂ = 2.20 s? Express your answer in meters per second. = 7.10 m/s Submit ✓ Correct Correct answer is shown. Your answer 6.78 m/s was either rounded differently or used a different number of significant figures than required for this part. Part B Previous Answers If the bus's position at time ₁ = 1.158 is 6.00 m, what is its position at time (₂2.20 s? Express your answer in meters. ΓΓΙ ΑΣΦ VO z= 7.90 A Submit Previous Answers Request Answer < Return to Assignment P * Incorrect; Try Again; 8 attempts remaining You may have omitted the initial velocity of the bus. Provide Feedback ? m3. In Physics, the height of a free-falling object at a given time t is given by h = –16t² + v,t + h., where v, is the initial velocity, and h, is the initial height. A student throws a ball upward from a height of 48ft, initially at 32 ft/s. What is the maximum height of the ball?A car starts from rest. It accelerates at constant acceleration for 5.0 seconds during which the car covers a distance of 125 m. After that the car continues at that speed for 22 seconds, after which the car slows down at a constant deceleration of – 4.5 m/sec2 until coming to a complete stop. Draw a v vs t, graph, label all the relevant data, and use the graphical method to determine total distance traveled.
- An object moves along the x-axis with its position x given as a function of time t by x (t) = Ct2 – Dt + B What is the object's velocity v as a function of time? v (t) =A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 6.25 m/s. The car is a distance d away. The bear is 21.3 m behind the tourist and running at 8.64 m/s. The tourist reaches the car safely. What is the maximum possible value for d? Vbear Number i Vtourist UnitsA golf ball is dropped from rest from a height of 9.90 m. It hits the pavement, then bounces back up, rising just 6.30 m before falling back down again. A boy then catches the ball when it is 1.40 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch. Number i Units
- Problem 3: You throw an object up with a speed of voy = 7 m/s from a height of y = 25 m. %3D Part (a) How long, in seconds, does it take for the object to reach the ground? Numeric : A numeric value is expected and not an expression. %3DA circular racetrack has a distance of 755 m. A racer is currently driving her car on the racetrack. As she starts the car from rest, it begins to accelerate to a speed of 120 kph in 9 s. a. Determine the magnitude of the total acceleration of the car, 4 s after the car begins to increase its speed. b. If she maintains the speed of the car to be 120 kph, determine the magnitude of the total acceleration of the car after 5s. c. Upon accelerating to 120 kph, will the car pass its starting point? Prove using kinematic equations.A car is moving with a velocity v(t)= 100-t2 in units of ft/sec where t is in seconds when the brakes are applied at t = 0. How far will the car travel before it comes to a stop, and what is its average velocity during that time?
- NASA has fired a rocket straight up into the air from ground level with an unknown initial velocity. It reaches a maximum height of 164,000 feet. This happened in Cape Canaveral therefore the rocket is under the acceleration of Earth's gavity in which we all know is -32 ft/sec/sec. Please round your answers to two decimal places. a). Find the time it takes for the rocket to reach maximum height. b). find the initial velocity necessary to reach that height.Velocity Problem 1. The position of an object is given as a function of time as x(t) = (3.0 m/s)t + (2.0 m/s2)t2. What is the average velocity of the object between t-0.0s and t = 2.0 SPan object has a velocity of (5.4 m/s)i - (5.8 m/s)j. over a period of 1.3 s, its velocity changes to (1.7 m/s)i + (5.9 m/s)j. What is the displacement