7) For parts a and b, assume that a jet accelerates from rest up to a speed of 80 m/s in a time of 1.8 seconds. a) What is the average acceleration of the jet during the 1.8 seconds? ) If the acceleration were constant (and equal to the above average) during the 1.8 seconds, (51 how far did the jet go during the 1.8 seconds? 5] . c) If the jet had an initial velocity of 15 m/s instead of O m/s, but still accelerated at the value in part (a) for 1.8 seconds, what would the final speed be after the 1.8 seconds?
Solution
Solution:
7) a) Intial vellocity = vi = 0
Final velocity = vf = 80 m/s
time = t = 1.8 s
Acceleration = a = (vf - vi) / t = (80 - 0 ) / 1.8 = 44.44 m/s 2 .
b) distance = d = 1/2 a t^2 = 1/2 * (44.44) (1.8)^2 = 72 m
c) vi = 15 m/s ;
vf = vi +at
=> Final velocity = vf = (15) + [(44.44)(1.8)] = 95 m/s .
.
1. 7) For parts a and b, assume that a jet accelerates from rest up to a speed of 80 m/s in a time of
1.8 seconds. a) What is the average acceleration of the jet during the 1.8 seconds? ) If the
acceleration were constant (and equal to the above average) during the 1.8 seconds, (51 how far
did the jet go during the 1.8 seconds? 5] . c) If the jet had an initial velocity of 15 m/s instead of
O m/s, but still accelerated at the value in part (a) for 1.8 seconds, what would the final speed be
after the 1.8 seconds?
Solution
Solution:
7) a) Intial vellocity = vi = 0
Final velocity = vf = 80 m/s
time = t = 1.8 s
Acceleration = a = (vf - vi) / t = (80 - 0 ) / 1.8 = 44.44 m/s 2 .
b) distance = d = 1/2 a t^2 = 1/2 * (44.44) (1.8)^2 = 72 m
c) vi = 15 m/s ;
vf = vi +at
=> Final velocity = vf = (15) + [(44.44)(1.8)] = 95 m/s .
![7) For parts a and b, assume that a jet accelerates from rest up to a speed of 80 m/s in a time of
1.8 seconds. a) What is the average acceleration of the jet during the 1.8 seconds? ) If the
acceleration were constant (and equal to the above average) during the 1.8 seconds, (51 how far
did the jet go during the 1.8 seconds? 5] . c) If the jet had an initial velocity of 15 m/s instead of
O m/s, but still accelerated at the value in part (a) for 1.8 seconds, what would the final speed be
after the 1.8 seconds?
Solution
Solution:
7) a) Intial vellocity = vi = 0
Final velocity = vf = 80 m/s
time = t = 1.8 s
Acceleration = a = (vf - vi) / t = (80 - 0 ) / 1.8 = 44.44 m/s 2 .
b) distance = d = 1/2 a t^2 = 1/2 * (44.44) (1.8)^2 = 72 m
c) vi = 15 m/s ;
vf = vi +at
=> Final velocity = vf = (15) + [(44.44)(1.8)] = 95 m/s .](https://image.slidesharecdn.com/7forpartsaandb-assumethatajetacceleratesfromrestuptoa-230127024337-fb54e437/85/7-for-parts-a-and-b-assume-that-a-jet-accelerates-from-rest-up-to-adocx-1-320.jpg)
