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In a test run, a certain car accelerates uniformly from zero to 20.4 m/s in 2.60 s.

Required:
a. What is the magnitude of the cars acceleration?
b. How long does it take the car to change speed from 10.0 m/s to 20 m/s.
c. Will doubling the time always double the change in speed? why?

Sagot :

CintiaSalazar

Answer:

(a) The acceleration is 7.85 m/s²

(b) It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.

(c) Doubling the time will double the change in velocity if the acceleration is kept constant.

Explanation:

(a)  Acceleration is the physical quantity that measures the rate of change of velocity with time. That is, acceleration relates changes in speed with the time in which they occur, that is, it measures how fast the changes in speed are.

The average acceleration is calculated using the following expression:

[tex]a=\frac{vf-vi}{t}[/tex]

where a is the acceleration, vf is the final velocity, vi is the initial velocity and t is the time.

In this case:

  • vf= 20.4 m/s
  • vi=0 m/s
  • t= 2.60 s

Replacing:

[tex]a=\frac{20.4 \frac{m}{s} - 0\frac{m}{s} }{2.60 s}[/tex]

a= 7.85 m/s²

The acceleration is 7.85 m/s²

(b) In this case you know:

  • a= 7.85 m/s²
  • vf= 20 m/s
  • vi= 10 m/s

Replacing:

[tex]7.85 \frac{m}{s^{2} } =\frac{20 \frac{m}{s} - 10\frac{m}{s} }{t}[/tex]

and solving you get:

[tex]t=\frac{20 \frac{m}{s} - 10\frac{m}{s} }{7.85 \frac{m}{s^{2} } }[/tex]

t=1.27 s

It takes the car to change speed from 10.0 m / s to 20 m / s in a time of 1.27 seconds.

(c)  Being:

[tex]a=\frac{vf-vi}{t}[/tex]

Then:

a*t= vf - vi

vf - vi represents the change in velocity. You can see that, if a (acceleration) is constant, then (vf - vi) is directly proportional to the time t: therefore, if t doubles, the change in velocity doubles as well.

In other words, doubling the time will double the change in velocity if the acceleration is kept constant.

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