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Artemon [7]
3 years ago
6

The graph above shows the position and time calculate the velocity of the particle from T=0s to T=4s?

Physics
1 answer:
zzz [600]3 years ago
3 0

Answer:

The velocity of the particle from T = 0 s to T = 4 s is;

0.5 m/s

Explanation:

The given parameters from the graph are;

The initial displacement (covered) at time, t₁ = 0 s is x₁ = 1 m

The displacement covered at time, t₂ = 4 s is x₂ = 3 m

The graph of distance to time, from time t = 0 to time t = 4 is a straight line graph, with the velocity given by the rate of change of the displacement to the time which is dx/dt which is also the slope of the graph given as follows;

The \ slope \ of \ the \ displacement \ time \ graph, \ m =velocity, \ v= \dfrac{x_2 - x_1}{t_2 - t_1}

Therefore, \ the \ velocity \ of \ the \ particle \ v  = \dfrac{3 \  m - 1 \ m}{4 \ s - 0 \ s}  = \dfrac{2 \ m}{4 \  s} = \dfrac{1}{2} \ m/s

The velocity of the particle from t = 0 s to t  = 4 s = 1/2 m/s = 0.5 m/s.

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daser333 [38]

Answer:

v = 98.75 km/h

Explanation:

Given,

The distance driver travels towards the east, d₁ = 135 km

The time period of the travel, t₁ = 1.5 h

The halting time, tₓ = 46 minutes

The distance driver travels towards the east, d₂ = 215 km

The time period of the travel, t₁ = 2 h

The average speed of the vehicle before stopping

                                    v₁ = d₁/t₁

                                        = 135/1.5

                                       = 90 km/h

The average speed of vehicle after stopping

                                    v₂ = d₂/t₂

                                         = 215/2

                                        = 107.5 km/h

The total average velocity of the driver

                                      v = (v₁ +v₂) /2

                                         = (90 + 107.5)/2

                                         = 98.75 km/h

Hence, the average velocity of the driver, v = 98.75 km/h

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Answer:

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Explanation:

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Answer:

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Explanation:

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