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mihalych1998 [28]

3 years ago

10

A motorist driving a 1000-kg car wishes to increase her speed from 20 m/s to 30 m/s in 5 s. Determine the horsepower required to

accomplish this increase. Neglect friction.

1 answer:

ikadub [295]3 years ago

8 0

Answer:

13.41 horsepower

Explanation:

Power: This can be defined as the rate at which energy is used up by a body. The S.I unit of power is Watt (W).

The expression for power is given as,

P = Ek/t...................... Equation 1

Where P = power, Ek = Kinetic Energy of the car, t = time.

Recall that,

Ek = 1/2mΔv²............ Equation 2

Where m = mass of the car, Δv = change in velocity of the car.

Substitute equation 2 into equation 1

P = 1/2(mΔv²)/t...................... Equation 3

Given:m = 1000 kg, Δv = 30-20 = 10 m/s, t = 5 s

Substitute into equation 3

P = 1/2(1000×10²)/5

P = 10000 W.

If,

745.7 W = 1 hp,

Then,

10000 W = (10000/745.7) hp

= 13.41 hp.

P = 13.41 horsepower

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Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

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Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

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