Acceleration is the magnitude of average velocity.
1 answer:
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Answer:
Δt =
Explanation:
Hi there!
Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.
The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):
v = d/t
Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:
Vr = D/ (2 · Δt1)
For the other half of the trip the expression of velocity will be:
Vw = D/(2 · Δt2)
The total time traveled is the sum of both Δt:
Δt(total) = Δt1 + Δt2
Then, solving the first equation for Δt1:
Vr = D/ (2 · Δt1)
Δt1 = D/(2 · Vr)
In the same way for the second equation:
Δt2 = D/(2 · Vw)
Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)
Δt(total) = D/2 · (1/Vr + 1/Vw)
The time needed by Rick to complete the trip was:
Δt(total) = D/2 · (1/Vr + 1/Vw)
Now let´s calculate the time it took Tim to do the trip:
Tim walks half of the time, then his speed could be expressed as follows:
Vw = 2d1/Δt Where d1 is the traveled distance.
Solving for d1:
Vw · Δt/2 = d1
He then ran half of the time:
Vr = 2d2/Δt
Solving for d2:
Vr · Δt/2 = d2
Since d1 + d2 = D, then:
Vw · Δt/2 + Vr · Δt/2 = D
Solving for Δt:
Δt (Vw/2 + Vr/2) = D
Δt = D / (Vw/2 + Vr/2)
Δt = D/ ((Vw + Vr)/2)
Δt = 2D / (Vw + Vr)
The time needed by Tim to complete the trip was:
Δt = 2D / (Vw + Vr)
Let´s find the diference between the time done by Tim and the one done by Rick:
Δt(tim) - Δt(rick)
2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))
= Δt
Let´s check the result. If Vr = Vw:
Δt = 2D/2Vr - D/2Vr - D/2Vr
Δt = D/Vr - D/Vr = 0
This makes sense because if both move with the same velocity all the time both will do the trip in the same time.
Answer:
Both of you did the same work but you expended more power.
Explanation:
<em>Work done</em> by an object is calculated by force applied multiplied by the distance.
W=F*d
From the figure given below let us calculate force applied bith you and yopur friend.
Let us take the stairs in positive x direction,
Work done by you W₁ ,
The force applied Fₓ = F - mgsinθ =maₓ
here aₓ = 0, because both of you move with constant speed
F - mgsinθ = 0
F= mgsinθ
The work done by you on the suitcase is
W = F L cos0° , where L is he length of the staircase.
W = FL = mgsinθL , by substituting value of F
Work done by you is W₁ = mgLsinθ
Similarly work done by your friend is W₂ = mgLsinθ.
Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .
Therefore <em>work done by both of you is same</em> . Both of you did equal work.
The power , is defined as amount of energy converted or transfered per second or rate at which work is done .
P = =
Power spend by you P₁ = mgLsinθ/t
P₁ = 15*9.8*Lsinθ/30
P₁ = 4.9L sinθ eqn 1
Power spend by your friend is P₂ = mgLsinθ/t
P₂ =15*9.8*Lsinθ/60
P₂ = 2.45Lsinθ eqn 2
Dividing eqn 1 and eqn 2
P₁ = 2P₂
You have spend more power than your friend .
Hence Both of you did equal work but you spend more power.
