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vampirchik [111]
3 years ago
13

A banked circular highway is designed for traffic moving

Physics
1 answer:
Nuetrik [128]3 years ago
3 0

Answer:

0.063

Explanation:

velocity of the car, v = 40 km/h = 11.11 m/s

radius, r = 200 m

Let the coefficient of friction is μ.

The coefficient of friction relates to the velocity on banked road is given by

\mu =\frac{v^{2}}{rg}

where, v is the velocity, r be the radius of the curve road and μ is coefficient of friction.

By substituting the values, we get

\mu =\frac{11.11^{2}}{200\times 9.8}

μ = 0.063

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a person using a machine applies a force of 100 newton's over a distance of 10 Meters to raise a 500 n object 1.5 meters what is
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To measure the strength of an earthquake, you can use either a _____ scale or _____ scale.
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Two vehicles A and B accelerate uniformly from rest.
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Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, a_A = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - a_A·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, V_{18A} = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, a_B = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ v_{18B} = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, v_{18B} = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle S_A = A₁ + A₂ + A₃

∴ S_A = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle S_A = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, S_B = 440 m

The distance of the two vehicles apart at the 18t second, S_{AB} = S_B - S_A

∴ S_{AB} = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, S_{AB} = 60 m.

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3 years ago
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