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Papessa [141]
3 years ago
12

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m

/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 6.76 s. How much time does the driver of the car measure for his trip between the poles
Physics
1 answer:
denis23 [38]3 years ago
8 0

Answer:

4.245s

Explanation:

Given that,

Hypothetical value of speed of light in a vacuum is 18 m/s

Speed of the car, 14 m/s

Time given is 6.76 s, and we're asked to find the observed time, T

The relationship between the two times can be given as

T = t / √[1 - (v²/c²)]

The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject

t = T / √[1 - (v²/c²)]

And now, we substitute the values and insert into the equation

t = 6.76 * √[1 - (14²/18²)]

t = 6.76 * √[1 - (196/324)]

t = 6.76 * √(1 - 0.605)

t = 6.76 * √0.395

t = 6.76 * 0.628

t = 4.245 s

Therefore, the time the driver measures for the trip is 4.245s

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2.85 A police car is traveling at a velocity of 18.0 m/s due north, when a car zooms by at a constant velocity of 42.0 m/s due n
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Answer:

11.1 s

Explanation:  

Speed of the police car as given = v = 18 m/s

Speed of the car = V = 42 m/s

Reaction time = t = 0.8 s

Distance traveled by the police car during the reaction time = d₁= 0.8 x 18 = 14.4 m

Distance traveled by speeding car = d₂ =0.8 x 42 = 33.6 m

Acceleration of the police car = a = 5 m/s/s

The police car can catch the speeding car only if it travels a distance equal to the speeding car in a time t.

Distance traveled by the police car = D = d₁ + v t +0.5 at², according to the kinematic equation.

⇒ D = 14.4 + 18 t + 0.5 (5) t²

⇒ D = 14.4 + 18 t+2.5 t²  → (1)

For the speeding car, distance traveled is D = 33.6 + 42 t, since it is constant velocity. Substitute for D from the above equation (1).

⇒ 14.4 + 18 t+2.5 t²=  33.6 + 42 t

⇒ 2.5 t² -24 t - 19.2 = 0

⇒ t = 10.3 s

Total time = t +0.8 s

⇒ Time taken for the police car to reach the speeding car = 10.3+0.8= 11.1 s

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