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FinnZ [79.3K]
3 years ago
10

In 1967, new zealander burt munro set the world record for an indian motorcycle, on the bonneville salt flats in utah, with a ma

ximum speed of 82.1 m/s. the one-way course was 8.045 km long. if it took burt 4.00 s to reach a velocity of 26.82 m/sec, how long (in seconds) did it take burt to reach his maximum speed? how far (in meters) did he travel during his acceleration? look at the equations in the last section. find acceleration first, then the time to accelerate to 82.1 m/s, then the x displacement during the time elapsed.

Physics
1 answer:
Fed [463]3 years ago
3 0
Draw a velocity-time diagram as shown below.

Because a velocity of 26.82 m/s is attained in 4.00 s from rest, the average acceleration is
a = 26.82/4 = 6.705 m/s²
The time required to reach maximum velocity of 82.1 m/s is
t₁ = (82.1 m/s)/(6.705 m/s²) = 12.2446 s

The distance traveled during the acceleration phase is
s₁ = (1/2)at₁²
    = (1/2)*(6.705 m/s²)*(12.2446 s)²
    = 502.64 m

Answer:
The time required to reach maximum speed is 12.245 s
The distance traveled during the acceleration phase is 502.6 m

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Complete Question

The diagram for this question is showed on the first uploaded image (reference homework solutions )

Answer:

The  velocity at the bottom is  v  = 11.76  \ m/ s

Explanation:

From the question we are told that

   The  total distance traveled is  d =  1.2  \ m

    The mass of the block is  m_b  =  0.3 \ kg

      The  height of the block from the ground is h =  0.60 m  

According the law of  energy  

   PE  =  KE

Where  PE  is the potential energy which is mathematically represented as

      PE  =  m * g  *  h

substituting values

     PE  =   3 *  9.8  *  0.60

      PE  =  17.64 \  J

So

   KE  is the kinetic energy at the bottom which is mathematically represented as

          KE  =  \frac{1}{2}  *  m v^2

So

      \frac{1}{2}  *  m* v ^2  =  PE

substituting values  

  =>    \frac{1}{2}  *  3 * v ^2  = 17.64

=>       v  = \sqrt{ \frac{ 17.64}{ 0.5 * 3 } }

=>    v  = 11.76  \ m/ s

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Newton’s cradle is a contraption where metal balls hang from a frame. When one ball is pulled and released, the collision causes
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An Object, Start from rest w Confront Aiceleration 8m/s2 along a
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Answer:

A)   v = 40 m / s, B)   v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

         

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

         v = vo + a t

         v = 0 + 8 5

         v = 40 m / s

B) the average velocity can be found with the relation

         v_average = vf + vo / 2

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At the end of a delivery ramp, a skid pad exerts a constant force on a package so that the package comes to rest in a distance d
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Answer:

increases by a factor of \sqrt{2}

Explanation:

First we need to find the initial velocity for it to stop at the distance 2d using the following equation of motion:

v^2 - v_0^2 = 2a\Delta s

where v = 0 m/s is the final velocity of the package when it stops, v_0 is the initial velocity of the package when it, a is the deceleration, and \Delta s = d is the distance traveled.

So the equation above can be simplified and plug in Δs = d, v_0 = v_1 for the 1st case

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let equation (2) divided by (1) we have:

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t = v / a

The time would increase by a factor of \sqrt{2}

7 0
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