Answer:
(a) [tex]v_f=28.5m/s[/tex]
(b) [tex]v_f=7.5m/s[/tex]
Explanation:
Hello.
(a) In this case since the car is moving at an initial velocity of 18 m/s due north, the final velocity is computed considering the acceleration as positive since it is due north as well:
[tex]v_f=v_0+at=18m/s+1.5m/s^2*7s\\\\v_f=28.5m/s[/tex]
(b) In this case, since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:
[tex]v_f=v_0+at=18m/s-1.5m/s^2*7s\\\\v_f=7.5m/s[/tex]
Best regards.
(a) The final velocity of car after 7.0 s for an acceleration is 1.5 m/s2 due north is 28.5 m/s.
(b) The final velocity of car after 7.0 s for an acceleration is 1.5 m/s2 due south is 7.5 m/s.
Given data:
The initial velocity of car due north is, u = 18 m/s.
The time interval is, t = 7.0 s.
(a)
For the acceleration of 1.5 m/s2 due north, the final velocity of car can be obtained from the first kinematic equation of motion as,
[tex]v = u +at[/tex]
Here, v is the magnitude of final velocity of car.
Solving as,
[tex]v = 18+(1.5)7\\\\v = 28.5 \;\rm m/s[/tex]
Thus, we can conclude that the final velocity of car after 7.0 s for an acceleration is 1.5 m/s2 due north is 28.5 m/s.
(b)
since the car is moving due north by the acceleration is due south it is undergoing a slowing down process, thereby the acceleration is negative therefore the final velocity turns out:
[tex]v'=u+(-a)t\\\\v'=18+(-1.5)7\\\\v' = 7.5 \;\rm m/s[/tex]
Thus, we can conclude that the final velocity of car after 7.0 s for an acceleration is 1.5 m/s2 due south is 7.5 m/s.
Learn more about the linear acceleration here:
https://brainly.com/question/14214161
