After locating a delicious bug, a hungry bat undergoes an acceleration of 4.5 m/s^2 (N 10 degrees E) for 0.78 seconds, emerging  from the turn with a velocity of 6.4m/s (S 35 degrees E). Determine its velocity before it began its turn.. Include a diagram.

Use the first equation of motion that is, v = u + at to find the initial velocity u of the bat.

Step-by-step explanation

N W < - E S Given : - a = 4.5 m s- 2 ( NIDE ) Jor O .78 8 . V = 6 . 4 ms ( 5 35 . E ) u = 8 using the first
equation of motion, ie. v = ut at we have u = v - at = 6. 4 - ( 4. 5 X 0-78 ) = 6- 4- 3.51 = 2 . 89 ms
Therefore the velocity before the bat began its turn is 2 . 89 m /s

Please see attached file