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An automobile and train move together along parallel paths at 23.2 m/s. The automobile then undergoes a uniform acceleration of −4 m/s 2 because of a red light and comes to rest. It remains at rest for 22.7 s, then accelerates back to a speed of 23.2 m/s at a rate of 2.09 m/s 2 . How far behind the train is the automobile when it reaches the speed of 23.2 m/s, assuming that the train speed has remained at 23.2 m/s? Answer in units of m.

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Answer:

Train will 588.215 m ahead of automobile

Explanation:

We have given initially speed of both train and automobile is same as 23 m/sec

After that acceleration of automobile [tex]a=-4m/sec^2[/tex]

Initial velocity u = 23 m/sec

Final velocity v = 0 ( as it comes to rest )

According to first law of motion

v=u+at

[tex]0=23-4\times t[/tex]

t = 5.75 sec

Now according to third law of motion [tex]v^2=u^2+2as[/tex]

[tex]0^2=23^2-2\times 4\times s[/tex]

s = 66.125 m

Now automobile accelerates at with acceleration [tex]a=2.09m/sec^2[/tex]

Initial velocity u = 23 m/sec

Final velocity v = 23.2 m/sec

So [tex]time\ t =\frac{v-u}{a}=\frac{23.2-23}{2.09}=0.0956sec[/tex]

Distance traveled by automobile in 0.0956 sec

[tex]s=ut+\frac{1}{2}at^2=23\times 0.0956+\frac{1}{2}\times 2.09\times 0.0956^2=2.208m[/tex]

So total distance traveled by automobile = 66.125+2.208 = 68.333 m

Total time = 5.75 +22.7+0.0956=28.545 sec

Sp distance traveled by train in 28.545 sec

= 28.545×23 = 656.548 m

So distance between automobile and train = 656.548 - 68.33=588.215 m

So train will 588.215 m ahead of automobile

The distance between the train and the automobile when it reaches the speed of 23.2 m/s is 196.05 m.

What time is taken by the car to stop?

To find the time taken by the car to stop, we will use the first equation of motion,

[tex]v-u =at\\\\[/tex]

since the final velocity of the car is zero because it's trying to stop.

[tex]v-u =at\\\\\dfrac{0-23.2}{-4}=t_1\\\\t_1 = 5.8\rm\ sec[/tex]

What is the time taken by the car to reach a velocity of 23.2 m/s?

To find the time taken by the car to reach the velocity, we will use the first equation of motion,

[tex]v-u =at\\\\[/tex]

since the initial velocity of the car is zero because it's was at rest,

[tex]v-u =at\\\\\dfrac{23.2-0}{2.09}=t_1\\\\t_1 = 11.1\rm\ sec[/tex]

What is the distance traveled by train during the entire journey?

We know that the velocity of the train is constant therefore, 23.2 m/s,

[tex]\rm Distance = speed \times time[/tex]

[tex]\rm Distance = 23.2\times (5.8+11.1)\\\\\rm Distance = 392.08\ meter[/tex]

What is the distance traveled by car?

The distance traveled by car in time([tex]t_1[/tex]),

[tex]S = ut+\dfrac{1}{2}at^2[/tex]

[tex]S_1 = (23.2\times 5.8)+\dfrac{1}{2}(-4\times 5.8^2)\\\\S_1 =134.56 - 67.28 \\\\S_1 = 67.28[/tex]

The distance traveled by car in time([tex]t_2[/tex]),

[tex]S_2 = (0 \times 11.1)+\dfrac{1}{2}(2.09\times 11.1^2)\\S_2 = 128.75 \rm \ m[/tex]

The total distance traveled by car,

[tex]S = S_1+S_2\\S = 196.03\rm\ m[/tex]

How far behind the train is the automobile when it reaches the speed of 23.2 m/s?

We know the distance traveled by train and the car,

[tex]\rm Distance\ between\ them = Distance_{Train} - S[/tex]

[tex]\rm Distance\ between\ them = 392.08-196.03 = 196.05\ m[/tex]

Hence, the distance between the train and the automobile when it reaches the speed of 23.2 m/s is 196.05 m.

Learn more about Equations of motion:

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