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gogolik [260]
3 years ago
13

A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

are both traveling at the same speed of 18.0 m/s, and the distance between them is 58.0 m . After t1 = 5.00 secs, the motorcycle starts to accelerate at a rate of 4.00 m/s^2 . The motorcycle catches up with the car at some time t2 . A) How long does it take from the moment when the motorcycle starts to accelerate until it catches up with the car? In other words, find t2-t1 Express the time numerically in seconds using three significant figures. B) How far does the motorcycle travel from the moment it starts to accelerate (at time t1 ) until it catches up with the car (at time t2)? Should you need to use an answer from a previous part, make sure you use the unrounded value. Answer numerically in meters using three significant figures.
Physics
1 answer:
xz_007 [3.2K]3 years ago
3 0

Answer:

a) \Delta{t} = 5.39s

b) the motorcycle travels 155 m

Explanation:

Let t_2-t_1 = \Delta{t}, then consider the equation of motion for the motorcycle (accelerated) and for the car (non accelerated):

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}}

where:

v_{m2} is the speed of the motorcycle at time 2

v_{c} is the velocity of the car (constant)

v_{0} 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:

\left[\begin{array}{cc}car&motorcycle\\x=v_0\Delta{t}&x+d=(\frac{v_0+v_{m2}}{2}}) \Delta{t}\end{array}\right]

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

For the second part, we need to calculate x+d, so you can use the equation of the car to calculate x:

x = v_0\Delta{t}= 18\sqrt{29}=96.933m\\then:\\x+d = 154.933

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