Question

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 28.0 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

Answer 1

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,

$F=ma$

This means that the acceleration is also constant.

Now we want to find how long the car takes to accelerate to a final velocity of

$v=57.0 mph \cdot \frac{1609}{3600}=25.5 m/s$

From an initial velocity of

u = 0

Using again the same suvat equation, and using the acceleration we found previously, we find:

$t=\frac{v-u}{a}=\frac{25.5-0}{8.9}=2.87 s$

Learn more about accelerated motion:

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About power:

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