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Westkost [7]
3 years ago
13

a driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di

splacement of 215 km, east. what is the driver’s average velocity?
Physics
1 answer:
daser333 [38]3 years ago
7 0

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:

The speed of the baseball is approximately 19.855 m/s

Explanation:

From the question, we have;

The frequency of the microwave beam emitted by the speed gun, f = 2.41 × 10¹⁰ Hz

The change in the frequency of the returning wave, Δf = +3190 Hz higher

The Doppler shift for the microwave frequency emitted by the speed gun which is then reflected back to the gun by the moving baseball is given by 2 shifts as follows;

 \dfrac{\Delta f}{f} = \dfrac{2 \cdot v_{baseball}}{c}

\therefore{\Delta f}{} = \dfrac{2 \cdot v_{baseball}}{c} \times f

Where;

Δf = The change in frequency observed, known as the beat frequency = 3190 Hz

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By plugging in the values, we have;

\therefore{\Delta f} = 3190 \ Hz =  \dfrac{2 \cdot v_{baseball}}{3.0 \times 10^8 \ m/s} \times 2.41 \times 10^{10} \ Hz

v_{baseball} = \dfrac{3190 \ Hz \times 3.0 \times 10^8 \ m/s }{2.41 \times 10^{10} \ Hz \times 2} \approx 19.855 \ m/s

The speed of the baseball, v_{baseball} ≈ 19.855 m/s

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Answer:

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Explanation:

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t=\frac{45 \ m}{12.5 \ m/s}

Divide. Note that the meters, or "m" will cancel each other out.

t=\frac{45 }{12.5 \ s}

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