Car A is accelerating in the direction of its motion at the rate of 3 ft /sec2. Car B is rounding a curve of 440-ft radius at a
constant speed of 30 mi /hr. Determine the velocity and acceleration which car B appears to have to an observer in car A if car A has reached a speed of 45 mi /hr for the positions represented.
1 answer:
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°
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<h2>Answer:</h2>
(a) v(t) = [10.0t - 12.0t²] m/s and a(t) = [10.0 - 24.0t ] m/s² respectively
(b) -28.0m/s and -38.0m/s² respectively
(c) 0.83s
(d) 0.83s
(e) x(t) = 1.1573 m [where t = 0.83s]
<h2>Explanation:</h2>The position equation is given by;
x(t) = 5.0t² - 4.0t³ m --------------------(i)
(a) Since velocity is the time rate of change of position, the velocity, v(t), of the particle as a function of time is calculated by finding the derivative of equation (i) as follows;
v(t) = dx(t) / dt = =
[ 5.0t² - 4.0t³ ]
v(t) = 10.0t - 12.0t² --------------------------------(ii)
Therefore, the velocity as a function of time is v(t) = 10.0t - 12.0t² m/s
Also, since acceleration is the time rate of change of velocity, the acceleration, a(t), of the particle as a function of time is calculated by finding the derivative of equation (ii) as follows;
a(t) = dx(t) / dt = =
[ 10.0t - 12.0t² ]
a(t) = 10.0 - 24.0t --------------------------------(iii)
Therefore, the acceleration as a function of time is a(t) = 10.0 - 24.0t m/s²
(b) To calculate the velocity at time t = 2.0s, substitute the value of t = 2.0 into equation (ii) as follows;
=> v(t) = 10.0t - 12.0t²
=> v(2.0) = 10.0(2) - 12.0(2)²
=> v(2.0) = 20.0 - 48.0
=> v(2.0) = -28.0m/s
Also, to calculate the acceleration at time t = 2.0s, substitute the value of t = 2.0 into equation (iii) as follows;
=> a(t) = 10.0 - 24.0t
=> a(2.0) = 10.0 - 24.0(2)
=> a(2.0) = 10.0 - 48.0
=> a(2.0) = -38.0 m/s²
Therefore, the velocity and acceleration at t = 2.0s are respectively -28.0m/s and -38.0m/s²
(c) The time at which the position is maximum is the time at which there is no change in position or the change in position is zero. i.e dx / dt = 0. It also means the time at which the velocity is zero. (since velocity is dx / dt)
Therefore, substitute v = 0 into equation (ii) and solve for t as follows;
=> v(t) = 10.0t - 12.0t²
=> 0 = 10.0t - 12.0t²
=> 0 = ( 10.0 - 12.0t ) t
=> t = 0 or 10.0 - 12.0t = 0
=> t = 0 or 10.0 = 12.0t
=> t = 0 or t = 10.0 / 12.0
=> t = 0 or t = 0.83s
At t=0 or t = 0.83s, the position of the particle will be maximum.
To get the more correct answer, substitute t = 0 and t = 0.83 into equation (i) as follows;
<em>Substitute t = 0 into equation (i)</em>
x(t) = 5.0(0)² - 4.0(0)³ = 0
At t = 0; x = 0
<em>Substitute t = 0.83s into equation (i)</em>
x(t) = 5.0(0.83)² - 4.0(0.83)³
x(t) = 5.0(0.6889) - 4.0(0.5718)
x(t) = 3.4445 - 2.2872
x(t) = 1.1573 m
At t = 0.83; x = 1.1573 m
Therefore, since the value of x at t = 0.83s is 1.1573m is greater than the value of x at t = 0 which is 0m, then the time at which the position is at maximum is 0.83s
(d) The velocity will be zero when the position is maximum. That means that, it will take the same time calculated in (c) above for the velocity to be zero. i.e t = 0.83s
(e) The maximum position function is found when t = 0.83s as shown in (c) above;
Substitute t = 0.83s into equation (i)
x(t) = 5.0(0.83)² - 4.0(0.83)³
x(t) = 5.0(0.6889) - 4.0(0.5718)
x(t) = 3.4445 - 2.2872
x(t) = 1.1573 m [where t = 0.83s]