Question

Light-rail passenger trains that provide transportation within and between cities speed up and slow down with a nearly constant (and quite modest) acceleration. A train travels through a congested part of town at 7.0 m/s . Once free of this area, it speeds up to 11 m/s in 8.0 s. At the edge of town, the driver again accelerates, with the same acceleration, for another 16 s to reach a higher cruising speed. You may want to review (Pages 42 - 45) .

Answer 1

19 m/s

Step-by-step explanation:

The complete question requires the final speed to be calculated.

Velocity is the rate and direction at which an object moves. Acceleration is the rate of change of velocity per unit time and can be calculated by the difference in velocity over a given time.

For this question, first the unknown acceleration must be calculated and used to determine the final velocity

Step 1: Calculate the acceleration

`a=\frac{v_(2)-v{1}}{t_(1)}`

`a=(11-7)/(8)`

`a=(4)/(8)`

`a=0.5 m/s^(2)`

Step 2: Calculate the velocity using the acceleration calculated above

`a=\frac{v_(3)-v{2}}{t_(2)}`

`0.5=(v_(3)-11)/(16)`

`v_(3)=19 m/s`