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3 years ago

What is the average acceleration of a car that starts from

Physics

2 answers:

MAXImum [283]3 years ago

4 0

Answer:

1.5m/s^2

Explanation:

In the picture above.

mr Goodwill [35]3 years ago

4 0

<u>Answer:</u>

The car is moving at an average acceleration of 1.5 m/s2

<u>Explanation</u>:

According to the question,

The car is moving in a straight road

The speed of the car = 18 m/s

Time is given (t) = 12 s

Acceleration is the rate of change of speed with respect to time. It is computed by the formula;

$\text { Acceleration }=\frac{\text {Speed }}{\text {Time}}$

Acceleration = (18 m/s)/(12 s)= 1.5 m/s2

$\text { Acceleration }=\frac{18 \mathrm{m} / \mathrm{s}}{12 \mathrm{s}}=1.5 \mathrm{m} / \mathrm{s}^{2}$

Therefore, the car is moving at an average acceleration of 1.5 m/s2

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3 years ago

A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f

Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

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N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

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