1) What is the distance of the Glider from the airport if it hasthe velocity of
ID: 1750676 • Letter: 1
Question
1) What is the distance of the Glider from the airport if it hasthe velocity of 180 km/hr, with an angle of 20 degrees and reachesa height of 2400m. Find the time, distance in the air, andacceleration.2) An airplane took off from the airport with a velocity of550 mi/hr, with an angle of 25 degrees. After 45 seconds it reachesits elevation. a) What is its height? b) How far is it from the airport with respect to theground?
3) The car is traveling with a speed of 80 mi/hr. It tries tostop for the red light and took 500 yards to stop. a) Find its acceleration b) How much time did it take to stop the car?
If you can give a step by step approach on how to solve theseproblems, that would really help. I'm just extremely confused onwhere to start off!
2) An airplane took off from the airport with a velocity of550 mi/hr, with an angle of 25 degrees. After 45 seconds it reachesits elevation. a) What is its height? b) How far is it from the airport with respect to theground?
3) The car is traveling with a speed of 80 mi/hr. It tries tostop for the red light and took 500 yards to stop. a) Find its acceleration b) How much time did it take to stop the car?
If you can give a step by step approach on how to solve theseproblems, that would really help. I'm just extremely confused onwhere to start off!
Explanation / Answer
Alright :) You should start by converting the units to S.I units(m,s,etc.) to get everything on the same terms. Then you canget right into it. Since the Glider takes off at a20o angle and the final height is 2400m, you canvisualize or draw a right triangle in which the height is the sideopposite of the angle. Thus you can use sin() =opposite/hypotenuse to solve for the hypotenuse which is the angleddistance traveled by the glider. hypotenuse =opposite/sin(). For the time (now that you have thedistance), velocity = distance/time, so time =distance/velocity. Plug in the numbers. Finally, sincethe velocity is given as a constant (180km/hr), what would theacceleration be? (hint. acceleration is the same as change invelocity) 1) Velocity unit change: 180 km/hr = 1.8x10^5m/hr = 500m/s Distance = 2400m/sin(20) ˜ 7017.13m (make sure calculator isis degree mode) Time = distance/velocity = 7017.13m/500m/s ˜ 14 seconds Acceleration = 0 m/s2 Hope this helps! I'll try to get to the others as soon aspossible
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