The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

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2 years ago

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The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

r (mph). At full power, the car can accelerate from zero to 28.0 mph in time 1.40 s.
The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hour (mph). At full power, the car can accelerate from zero to 28.0mph in time 1.40s .
Part A. At full power, how long would it take for the car to accelerate from 0 to 57.0mph ? Neglect friction and air resistance. THE ANSWER IS "6 seconds" according the mastering physics..
Part B. A more realistic car would cause the wheels to spin in a manner that would result in the ground pushing it forward with a constant force (in contrast to the constant power in Part A). If such a sports car went from zero to 28.0mph in time 1.40s , how long would it take to go from zero to 57.0mph ?
Express your answer numerically, in seconds

Physics

1 answer:

Mama L [17] · Answer 1 · 2023-04-25T19:24:49+03:00

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

$P=\frac{\frac{1}{2}mv^2}{t}=const.$

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

$\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}$

where:

$t_1=1.40 s$ is the time the car needs to accelerates to $v_1=28.0 mph$

$t_2$ is the time the car needs to accelerate to $v_2=57.0 mph$

Therefore, solving for $t_2$ ,

$t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s$

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

$a=\frac{v-u}{t}$

where:

u = 0 is the initial velocity

$v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s$ is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

$a=\frac{12.5-0}{1.40}=8.9 m/s^2$

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

$F=ma$