Answer:
26.4 m/s²
Explanation:
You want the magnitude of the acceleration that changes a velocity of 2.11 m/s at 37.0° to 3.80 m/s at 150° in 0.19 seconds.
Acceleration
The acceleration is the change in velocity divided by the time.
a = (3.80∠150° -2.11∠37.0° m/s)/0.19 s = 26.4∠173° m/s²
The magnitude of the acceleration is about 26.4 m/s².
__
Additional comment
The magnitude of the velocity difference can also be found using the law of cosines:
c² = a² +b² -2ab·cos(C)
c = √(a² +b² -2ab·cos(C)) = √(3.80² +2.11² -2(3.80)(2.11)cos(150°-37·))
c ≈ √25.1578 ≈ 5.0158 . . . . . |∆v|
Then the magnitude of the acceleration is ...
(5.0158 m/s)/(0.19 s) ≈ 26.4 m/s² . . . . . |a| = |∆v|÷t
The angle of the acceleration, relative to one of the velocity vectors, can be found using the law of sines.
