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Montano1993 [528]
3 years ago
5

- What is the total distance traveled by the car

Physics
1 answer:
slega [8]3 years ago
7 0

*Graph is attached with the answer*

Answer:

Total distance traveled by the car is 40 m

Explanation:

It can be seen from the graph that for the first 4 second car is moving with uniform acceleration and for next 2 second it has constant velocity

So, we solved this question in 2 parts

Part 1:

               initial velocity = v₁ = 0 m/s

               final velocity = v₂ = 10 m/s

                time = t = 4 s

                acceleration = a = (v₂ - v₁)/t

                                a = 10/4 = 2.5 m/s²

Using 3rd equation of motion

                2as = v₂² - v₁²

                 2(2.5)s = 10²

                      s₁ = 100/5

                     s₁ = 20 m

Part 2:

            s₂ = v₂t

            s₂ = 10 × 2

            s₂ = 20 m

Total distance = S = s₁ + s₂

                              = 20 + 20

                          S = 40 m

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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri
Ainat [17]

Answer:

The taken is  t_A  = 19.0 \ s

Explanation:

Frm the question we are told that

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     The distance of car B  from A is  d = 300 \ m

     The acceleration of car A is  a_A  = 2.40 \ m/s^2

For A to overtake B

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Now the this distance traveled by car B before it is overtaken by A is  

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Where t_B is the time taken by car B

Now this can also be represented as using equation of motion as

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Now substituting values

       d = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

Equating the both d

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substituting values

   29 * t_A = 22t_A  + \frac{1}{2} (2.40)^2 t_A^2 - 300

   7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300

  7 t_A =1.2 t_A^2 - 300

   1.2 t_A^2 - 7 t_A - 300  = 0

Solving this using quadratic formula we have that

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7 0
3 years ago
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Answer:

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SpyIntel [72]

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