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3 years ago

A man is driving his 1100 kg car at 36 km/h on a straight freeway. After accelerating for 30 seconds, the car

Physics

1 answer:

rusak2 [61]3 years ago

5 0

Answer:

W = 439998 J = 439.99 KJ

Explanation:

First, we will calculate the acceleration of the car by using the first equation of motion:

$v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}$

where,

a = acceleration = ?

vf = final speed = $108(\frac{km}{h})(\frac{1000\ m}{1\ km})(\frac{1\ h}{3600\ s})$ = 30 m/s

vi = initial speed = $36(\frac{km}{h})(\frac{1000\ m}{1\ km})(\frac{1\ h}{3600\ s})$ = 10 m/s

t = time = 30 s

Therefore,

$a = \frac{30\ m/s - 10\ m/s}{30\ s}$

a = 0.67 m/s²

Now, we will calculate the force applied by the engine:

F = ma

where,

F = force = ?

m = mass = 1100 kg

Therefore,

F = (1100 kg)(0.67 m/s²)

F = 733.3 N

Now, we will calculate the distance covered by the car by using the second equation of motion:

$s = v_it+\frac{1}{2}at^2\\\\s = (10\ m/s)(30\ s)+\frac{1}{2} (0.67\ m/s^2)(30\ s)^2$

s = 600 m

Now, the work done (W) by engine can be calculated as follows:

W = Fs

W = (733.3 N)(600 m)

W = 439998 J = 439.99 KJ

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6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri

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Answer:

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Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

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a = v² - u²/2s

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Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

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t = (0 m/s - 22.35 m/s)/-2.13 m/s²

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b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

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substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

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So, t = (v - u)/a

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