Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
The answer to this question is Initial velocity of plane will be 34236 m/s and 92437.2 Km is the distance travelled by it.
Three equation of motion are:-
- v = u + at
- s = ut + (1/2)at²
- v² - u² = 2as
Where v is final velocity, u in initial velocity, s is the displacement by the object, a is the acceleration and t denotes the time.
In question we have given deceleration as 6.34 m/s² and time as 1.5 hour which is equal to 5400 seconds.
Applying equation 1 to find the initial speed of plane
v = u + at
0 = u + (-6.34 × 5400) {v=0 as plane will stop after 5400 sec}
u = 6.34 × 5400
u = 34236 m/s
Initial velocity of plane is 34236 m/s
Applying equation 2 to find the displacement of plane in that time period
s = ut + (1/2)at²
s = ( 34236 × 5400 ) - ( (1/2) × 6.34 × 5400² )
s = 5400 × ( 34236 - ((1/2) × 6.34 × 5400) )
s = 5400 × ( 34236 - 17118 )
s = 5400 × 17118 metres
s = 5.4 × 17118 Km
s = 92437.2 Km
Distance travelled by plane is 92437.2 Km
So, the initial velocity of plane will be 34236 m/s and the displacement of plane in that time period will be 92437.2 Km.
Learn more about Motion here:
https://brainly.com/question/25951773
#SPJ10
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
