Question

A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous velocity is 16 m/s, 45° north of west. What is the magnitude of the average acceleration of the car during the four-second interval?

Answer 1

Answer:

$4.25 m/s^{2}$

Explanation:

Change in velocity considering the x component will be

Final velocity-Initial velocity

$\triangle v_x= 16cos 45^{\circ}-24=-12.6862915 m/s$

Change in velocity considering the y component will be

Final velocity-Initial velocity

$\triangle v_y= 16sin 45^{\circ}-0=11.3137085 m/s$

Resultant change in velocity$=\sqrt {(-12.6862915 m/s)^{2}+(11.3137085 m/s)^{2}}=16.9982938 m/s$

Acceleration= change in velocity per unit time hence

$a= \frac {16.9982938}{4}=4.24957345\approx 4.25 m/s^{2}$