Question

During 9.97 seconds a motorcyclist changes his velocity from v1,x = -36.5 m/s and v1,y = 10.9 m/s to v2,x =-21.9 m/s and v2,y = 27.5 m/s. Find the components of the motorcycle\'s average acceleration during this process, aav,x and aav,y.

Answer 1

Average acceleration is defined as

$a = \frac{v_f - v_i}{t}$

here

$v_f = final\: speed$

$v_i = initial\: speed$

$t = time$

now for x direction we can say

$a_x = \frac{v_{2x} - v_{1x}}{t}$

$a_x = \frac{-21.9 + 36.5}{9.97}$

$a_x = 1.46 m/s^2$

Similarly for Y direction

$a_y = \frac{v_{2y} - v_{1y}}{t}$

$a_y = \frac{27.5 - 10.9}{9.97}$

$a_y = 1.66 m/s^2$

so above is the acceleration in x and y directions