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jolli1 [7]
3 years ago
12

To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceler

ation. Suppose that the total distance moved by your car during these two phases is 56.9 m when its initial speed is 80.0 km/h, and 25.7 m when its initial speed is 50.7 km/h. What are (a) your reaction time and (b) the magnitude of the deceleration?
Physics
1 answer:
ANEK [815]3 years ago
8 0

Answer:

a) t_r = 0.55 s

b) a = 5.59 m/s²

Explanation:

given,

total distance traveled by the car to stop is 56.9 m when speed of vehicle is 80 km/h or 80 × 0.278 = 22.24 m/s

total distance traveled by the car to stop is 25.7 m when speed of vehicle is 50.7 km/h or 50.7 × 0.278 = 14.09 m/s

using stopping distance formula

s_1 = v_1 t_r +\dfrac{v_1^2}{2 a}................(1)

s_2 = v_2 t_r +\dfrac{v_2^2}{2 a}..............(2)

on solving both the equation we get

a = \dfarc{v_1v_2(v_1-v_2)}{2(s_1v_2-s_2v_1)}

a = \dfarc{22.4\times 14.09(22.24-14.09)}{2(56.9\times 14.09-25.7\times 22.24)}

a = 5.59 m/s²

now reaction time calculation

t_r =\dfrac{v_1^2d_2-v_2^2d_1}{v_1v_2(v_1-v_2)}

t_r =\dfrac{22.24^2\times 25.7-14.09^2\times 56.9}{22.4\times 14.09(22.24-14.09)}

t_r = 0.55 s      

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3 years ago
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A car going at v = 29.7 m/s (67 mph) rounds a curve of radius R = 50.0 m, where the road is banked at an angle of θ = 30.0°. Wha
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Answer:

μ = 0.6

Explanation:

given,

speed of car = 29.7 m/s

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θ = 30.0°

minimum static friction = ?

now,

writing all the forces acting along y-direction

N cos θ - f sinθ = mg

N cos θ -μN sinθ = mg

N = \dfrac{m g}{cos\theta-\mu sin \theta}

now, writing the forces acting along x- direction

N sin θ + f cos θ = F_{net}

N cos θ + μN sinθ = F_{net}

\dfrac{m g}{cos\theta-\mu sin \theta}(cos \theta + \mu sin\theta)=F_{net}

taking cos θ from nominator and denominator

F_{net} =\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{mv^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}. mg

\dfrac{v^2}{r}=\dfrac{tan\theta + \mu}{1-\mutan\theta}g

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\mu=\dfrac{29.7^2 -50 \times 9.8tan 30^0}{29.7^2\times tan 30^0 +50 \times 9.8}

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Being physically active helps in maintaining fat levels in the body of an individual, and it is healthy for the individual.

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