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
You would be 127.86 lbs. approx. One kilogram is equal to 2.20 pounds, so you just multiply kilos into the conversion rate. If you are trying to convert pounds to kilos, divide your pounds into the conversion rate.
Hope that helped! :)
B. True give me branliest