The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
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C. when you can't achieve your goal due to events beyond your control
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