Ahsoka pulls a wounded soldier (mass 64 kg) across the ground with a force 18 Newtons. If they have an acceleration of 1.5 m/s2
2 answers:
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Answer:
Incomplete question
Check attachment for the diagram of the problem.
Explanation:
The acceleration of the car A is given as
a=3ft/s²
Car B is rounding a curve of radius
r=440ft
Car B is moving at constant speed of Vb=30mi/hr.
Car A reach a speed of 45mi/hr
Note, 1 mile = 5280ft
And 1 hour= 3600s
Then
Va=45mi/hr=45×5280/3600
Va=66ft/s
Also,
Vb=30mi/hour=30×5280/3600
Vb=44ft/s
Now,
a. Let write the relative velocity of car B, relative to car A
Vb = Va + Vb/a
Then,
Using triangle rule, because vectors cannot be added automatically
Vb/a²= Vb²+Va²-2Va•VbCosθ
From the given graphical question the angle between Va and Vb is 60°.
Vb/a²=44²+66² - 2•44•66Cos60
Vb/a²=1936+ 4356 - 5808Cos60
Vb/a² = 3388
Vb/a = √3388
Vb/a = 58.21 ft/s
The direction is given as
Using Sine Rule
a/SinA = b/SinB = c/SinC
i.e.
Va/SinA = Vb/SinB = (Vb/a)/SinC
66/SinA = 44/SinB = 58.21/Sin60
Then, to get B
44/SinB = 58.21/Sin60
44Sin60/58.21 = SinB
0.6546 = SinB
B=arcsin(0.6546)
B=40.89°
b. The acceleration of Car B due to Car A.
Let write the relative acceleration of car B, relative to car A.
Let Aa be acceleration of car A
Ab be the acceleration of car B.
Ab = Aa + Ab/a
Given the acceleration of car A
Aa=3ft/s²
Then to get the acceleration of car B, using the tangential acceleration formular
a = v²/r
Ab = Vb²/r
Ab = 44²/440
Ab = 4.4ft/s²
Using cosine rule again as above
Ab/a²= Aa²+Ab² - 2•Aa•Ab•Cosθ
Ab/a²= 3²+4.4²- 2•3•4.4•Cos30
Ab/a²= 9+19.36 - 22.863
Ab/a² = 5.497
Ab/a = √5.497
Ab/a = 2.34ft/s²
To get the direction using Sine rule again, as done above
Using Sine Rule
a/SinA = b/SinB = c/SinC
i.e.
Aa/SinA = Ab/SinB = (Ab/a)/SinC
3/SinA = 4.4/SinB = 2.34/Sin30
Then, to get B
4.4/SinB = 2.34/Sin30
4.4Sin30/2.34 = SinB
0.9402 = SinB
B=arcsin(0.9402)
B=70.1°
Since B is obtuse, the other solution for Sine is given as
B= nπ - θ. , when n=1
B=180-70.1
B=109.92°
