Respuesta :
Answer:
a) [tex]\Delta{t} = 5.39s[/tex]
b) the motorcycle travels 155 m
Explanation:
Let [tex]t_2-t_1 = \Delta{t}[/tex], then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):
[tex]v_{m2}=v_0+a\Delta{t}\\x+d=(\frac{v_0+v_{m2}}{2} )\Delta{t}\\v_c = v_0 = \frac{x}{\Delta{t}}[/tex]
where:
[tex]v_{m2}[/tex] is the speed of the motorcycle at time 2
[tex]v_{c}[/tex] is the velocity of the car (constant)
[tex]v_{0}[/tex] is the velocity of the car and the motorcycle at time 1
d is the distance between the car and the motorcycle at time 1
x is the distance traveled by the car between time 1 and time 2
Solving the system of equations:
[tex]\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right][/tex]
[tex]v_0\Delta{t}=\frac{v_0+v_{m2}}{2}\Delta{t}-d \\\frac{v_0+v_{m2}}{2}\Delta{t}-v_0\Delta{t}=d\\(v_0+v_{m2})\Delta{t}-2v_0\Delta{t}=2d\\(v_0+v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\(2v_0+a\Delta{t})\Delta{t}-2v_0\Delta{t}=2d\\a\Delta{t}^2=2d\\\Delta{t}=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2*58}{4}}=\sqrt{29}=5.385s[/tex]
For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:
[tex]x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933[/tex]
The equations of motion are used to find the motion characteristics of an object with time
A) The time it takes the motorcycle to reach the car is approximately 5.39 seconds
B) The distance the motorcycle travel from the moment it starts to accelerate to the moment it catches up with the car is approximately 155 meters
The reason the above values are correct is as follows:
The given parameters are;
The initial speed of the car and the motorcycle, v₁ = 18.0 m/s
The initial distance between the motorcycle following the car = 58.0 m
The time after which the motorcycle starts to accelerate, t₁ = 5.00 secs
The acceleration of the motorcycle = 4.00 m/s²
The time at which the motorcycle catches up with the car = t₂
Required:
A) The time it takes the motorcycle to accelerate before it catches up with the car, which is to find t₂ - t₁
Method:
The distance moved by motorcycle to reach the car = The distance moved by the car in the same time + 58.0 meters
Solution;
The appropriate equation of motion to use are [tex]s = u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2[/tex] and [tex]d = u\cdot t[/tex]
Where;
s = The distance between the motorcycle and the car = 58.0 m
u = The initial velocity of the motorcycle = 18.0 m/s
t = Δt = t₂ - t₁ = The time the motorcycle accelerates
a = The acceleration of the motorcycle = 4.00 m/s²
d = The distance moved by the car
By the method, we have;
[tex]s = d + 58.0[/tex]
[tex]u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2 = u \cdot t[/tex]
Plugging in the values into the equation gives;
[tex](18.0 \times t) + \dfrac{1}{2} \times 4 \times t^2 = (18.0 \times t) + 58.0[/tex]
Cancelling the like term, (18.0 × t), on both sides of the equation gives;
[tex]\dfrac{1}{2} \times 4 \times t^2 = 58.0[/tex]
[tex]2\cdot t^2 = 58.0[/tex]
[tex]t = \Delta t = \sqrt{\dfrac{58.0}{2} } = \sqrt{29} \approx 5.39[/tex]
The time it takes the motorcycle to reach the car, t ≈ 5.39 seconds
The time it takes the motorcycle to reach the car is approximately 5.39 seconds
B) Required:
The distance the motorcycle travels from the moment it starts to accelerate, t₁ to the time it catches up with the car, t₂
Method:
The distance travelled during the time interval t₂ - t₁ should be calculated
Solution:
The distance, s, travelled during the time interval [tex]t_2 - t_1 = \Delta t = t[/tex] is given as follows;
[tex]s = u \cdot t + \dfrac{1}{2} \cdot a \cdot t^2[/tex]
From part (A), u = 18.0 m/s t = [tex]\sqrt{29}[/tex] s, and a = 4.00 m/s², therefore;
[tex]s = 18.0 \times \sqrt{29} + \dfrac{1}{2} \times 4.00 \times (\sqrt{29} )^2 \approx 154.933[/tex]
Rounded to three significant figures, the distance the motorcycle travel, s = 155 meters
Learn more about kinetic equation motion here:
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