Answer:
37.125 m
Explanation:
Using the equation of motion
s=ut+0.5at^{2} where s is distance, u is initial velocity, t is time and a is acceleration
<u>Distance during acceleration</u>
Acceleration, a=\frac {V_{final}-V_{initial}}{t} where V_{final} is final velocity and V_{initial} is initial velocity.
Substituting 0.0 m/s for initial velocity and 4.5 m/s for final velocity, acceleration will be
a=\frac {4.5 m/s-0 m/s}{4.5 s}=1 m/s^{2}
Then substituting u for 0 m/s, t for 4.5 s and a for 1 m/s^{2} into the equation of motion
s=0*4.5+ 0.5*1*4.5^{2}=0+10.125
=10.125 m
<u>Distance at a constant speed</u>
At a constant speed, there's no acceleration and since speed=distance/time then distance is speed*time
Distance=4.5 m/s*6 s=27 m
<u>Total distance</u>
Total=27+10.125=37.125 m
Answer:
4.02 km/hr
Explanation:
5 km/hr = 1.39 m/s
The swimmer's speed relative to the ground must have the same direction as line AC.
The vertical component of the velocity is:
uᵧ = us cos 45
uᵧ = √2/2 us
The horizontal component of the velocity is:
uₓ = 1.39 − us sin 45
uₓ = 1.39 − √2/2 us
Writing a proportion:
uₓ / uᵧ = 121 / 159
(1.39 − √2/2 us) / (√2/2 us) = 121 / 159
Cross multiply and solve:
159 (1.39 − √2/2 us) = 121 (√2/2 us)
220.8 − 79.5√2 us = 60.5√2 us
220.8 = 140√2 us
us = 1.115
The swimmer's speed is 1.115 m/s, or 4.02 km/hr.
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, 
= 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
Answer: 8 meters per second
Explanation: If you add 60 to 20 you get 80 meters and since he ran those 80 meters in 10 seconds you divide 80 by ten and get 8 and then you get 8m/s
