The Sun is to a planet as -
1 answer:
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Answer:
travel time is 32.4 s
maximum speed is 45.36 m/s
Explanation:
given data
distance = 980 m
acceleration = 4.20 m/s² for first 1/4 of that distance
acceleration = -1.40 m/s² for next 3/4 of that distance
to find out
travel time through the 980 m and maximum speed
solution
we know for first 1/4 of that distance is = = 245 m
so equation of motion
s = ut + 0.5 ×at² .............1
here u is initial speed = 0 and a is acceleration an t is time
s = ut + 0.5 ×at²
245 = 0+ 0.5 ×4.20 (t)²
t = 10.80 s
so
maximum speed at 1/4 of that distance
use equation of motion
v² - u² = 2as
put here value
v² - 0 = 2(4.20)× (245)
v = 45.36 m/s
so maximum speed is 45.36 m/s
and
for 3/4 distance
use equation of motion
v = u + at
here u is here 45.36 and a is acceleration and t is time and v final speed is 0
0 = 45.36 + (-1.40) × t
t = 32.4 s
so travel time is 32.4 s
Answer: Got It!
<em>Explanation: </em>let s = speed at launch
v = 0 at top = s sin 63 - g t
so at top
t = s sin 63/g = .0909 s
h = 13.6 = s sin 63 t - 4.9 t^2
13.6 = .081s^2 - .0405 s^2
s^2 = 336
s = 18.3 m/s
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