Answer:
Average speed = distance/time
From 1 to 9 seconds:
Distance covered = 1 - 0.2 = 0.8 km
Time = 9 - 1 = 8 sec
Average speed = 0.8 km / 8 sec
Average speed = 0.1 km/s .
The average speed for the whole test is 1.6 km / 20 sec = 0.08 km/sec. A graph of speed vs time would average out as a horizontal line at 0.08 km/sec from 1 sec to 21 sec. The area under it would be (0.08 km/s) x (20 sec) = 1.6 km.
Surprise surprise ! The area under a speed/time graph is the distance covered during that time !
In closing, I want to express my gratitude for the gracious bounty of 3 points with which I have been showered. Moreover, the green breadcrust and tepid cloudy water have also been refreshing.
Explanation:
<span>here's a cheap trick
it would take the same time to accelerate from rest to top speed
as it would take to decelerate from top speed to zero
so
instead of
d = Vi t + 1/2 a t^2 where Vi is positive and a is negative
we'll use
Vi = 0 and a is positive
giving
85 = 0 + 1/2 (0.43) t^2 = 0.215 t^2
t^2 = 395.345
t = 19.88s or 20. s to 2 sig figs
or we ccould find Vi from
Vf*2 = Vi^2 + 2 a d
0 = Vi^2 + 2 (0.43) 85
Vi^2 = 71.4
Vi = 8.45m/s
then
85 = 8.45 t + 1/2 (-0.43) t^2
85 = 8.45 t - 0.215 t^2
0.215 t^2 - 8.45 + 85 = 0
t = 19.65s or 20. s to 2 s.f.(minor difference arises from rounding Vi)
or another cheap trick
when a is constant
Vavg = (Vf + Vi) /2 = 8.45/2 = 4.225
and
d = Vavg t
85 = 4.225 t
t = 20.12 or 20. s to 2 s.f. (minor differences from intermidiate roundings)
anyway you choose you get 20. s</span>
Answer:
Explanation:
Given
Initial Velocity (u)=13 m/s
angle
distance between fence and deer=2.5 m
We consider deer jump similar to projectile motion
equation of trajectory
Thus deer will cross the fence with an difference in its jump and fence
1.5-1.148=0.351 m
so deer rises during when it is over fence
The answer is B.) I took the test.
Answer:
i need some more coins LMAI
