A train, traveling at a constant speed of 25.8 m/s, comes to an incline with a constant slope. While going up the incline, the train slows down with a constant acceleration of magnitude 1.66 m/s². What is the speed of the train after 8.30 s on the incline?

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

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.

double the distance = divide strength by 4

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s