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tamaranim1 [39]
3 years ago
11

An electric vehicle starts from rest and accelerates at a rate of 2.4 m/s2 in a straight line until it reaches a speed of 27 m/s

. The vehicle then slows at a constant rate of 1.3 m/s2 until it stops. (a) How much time elapses from start to stop? (b) How far does the vehicle move from start to stop?
Physics
1 answer:
DENIUS [597]3 years ago
6 0
A) use v=u+at for both

First section, v=27, u=0, a=2.4. You should get 11seconds.
Second section, v=0, u=27, a=-1.3. You should get 21seconds.
This means that the total time is 22seconds.

b) You can either use s=ut+0.5at^2 or v^2=u^2+2as. Personally, I would use the second one as you are not relying on your previous answer.

First section, v=27, u=0, a=2.4. You should get 152m.
Second section, v=0, u=27, a=-1.3. You should get 280m.
This makes your overall displacement 432m.
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Answer:

The maximum length during the motion is L_{max} = 1.45m

Explanation:

From the question we are told that

           The mass  is  m =0.5 kg

            The vertical spring  length is  L = 1.10m

            The unstretched  length is  L_{un} = 1.30m

          The initial speed is v_i = 1.3m/s

          The new length of the spring L_{new} =  1.30 m

The spring constant k is mathematically represented as

                           k = -\frac{F}{y}

Where F is the force applied  = m * g = 0.5 * 9.8=4.9N

           y is the difference in weight which is   =1.10-0.50=0.6m

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

    Now  substituting values accordingly

                    k =  \frac{4.9}{0.6}

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The  elastic potential energy is given as E_{PE} = \frac{1}{2} k D^2

  where D is this the is the displacement  

Since Energy is conserved the total elastic potential energy would be

             E_T = initial  \ elastic\ potential \ energy + kinetic \ energy

            E_T = \frac{1}{2} k D_{max}^2 =   \frac{1}{2} k D^2 + \frac{1}{2} mv^2

Substituting value accordingly

                \frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2

                4.085 * D_{max}^2 = 3.69

                 D^2_{max} = 0.9033

                D_{max} = 0.950m

So to obtain total length we would add the unstretched length

 So we have

                  L_{max} = 0.950 + 0.5 = 1.45m

                               

               

               

                 

                     

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