a car travels along a highway with a velocity of 24m/s, west. then exits the highway and 4s later, its instantaneous velocity is 16m/s, 45degrees north of west. Calculate the magnitude of the average accelaration of the car during the 5s interval?

Answer 1

i = unit vector x (east)
j = unit vector y (north)
Vo = -24 i + 0 j
Vf = -16/sqrt2 i + 16/sqrt2 j
change in velocity =(24-16/sqrt2)i + 16/sqrt2 j
dV/dt = A = (1/5)[(24-16/sqrt2)i + 16/sqrt2 j ]

To get the magnitude of A take the square root of the sum of the squares.