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rodikova [14]
3 years ago
8

A high-performance sports car can go from 0 to 100 mph in 5.8 s. (Assume the car travels in the positive direction. Indicate the

direction with the sign of your answer.) (a) What is the car's average acceleration (in SI units; meter-kilogram-second)The same car can come to a complete stop from 34 m/s in 5.1 s. What is its average acceleration?
Physics
1 answer:
sasho [114]3 years ago
7 0

Answer: a) 7.71 m/s², b) - 6.67 m/s²

Explanation: first thing to note is that

1 mile = 1609.34

1 hour = 3600s

Hence, 100mph to m/s = (100 × 1609.34)/3600 = 44.71 m/s

Initial velocity (u) = 0, final velocity (v) = 44.71 m/s, t = 5.8s, a = acceleration =?

By using newton's laws of motion

v = u + at

44.71 = 0 + a(5.8)

44.71 = 5.8a

a = 44.71/5.8

a = 7.71 m/s²

Question b)

The car is completing a stop which implies that the car is coming to rest, and when a car is coming to rest, the final velocity (v) is zero.

Hence u = 34 m/s, v = 0, a =?, t = 5.1 s

v = u + at

0 = 34 + a(5.1)

a(5.1) = - 34

a = - 34/5.1

a = - 6.67 m/s².

The negative sign beside the acceleration shows that the body is decelerating

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If you take the given information and try to apply wave motion to it:

             Wave speed = (wavelength) x (frequency)

             Frequency  =  (speed) / (wavelength) ,

you would end up with

             Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz

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This math is not applicable to the pendulum.

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On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
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Answer:

a)

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        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

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