A car traveling along the straight portions of the road has the velocities indicated in the figure when it arrives at points A,B, and C. If it takes 3 s to go from A to B, and then 5 s to go from B to C, determine the average acceleration between points A and B and between points A and C.

#### Solution:

We will first write the velocity of the car at locations, A, B and C in Cartesian vector form like so:

v_A=\left\{20i\right\} m/s

v_B=30\cos45^0i+30\sin45^0j=\left\{21.2i+21.2j\right\} m/s

v_C=\left\{40i\right\} m/s

v_B=30\cos45^0i+30\sin45^0j=\left\{21.2i+21.2j\right\} m/s

v_C=\left\{40i\right\} m/s

The average acceleration can be found by dividing velocity by time:

a=\dfrac{\Delta v}{\Delta t}

a_{AB}=\dfrac{(21.2i+21.2j)-(20i)}{3}

a_{AB}=\left\{0.4i+7.07j\right\} m/s^2

a_{AB}=\dfrac{(21.2i+21.2j)-(20i)}{3}

a_{AB}=\left\{0.4i+7.07j\right\} m/s^2

a_{AC}=\dfrac{40i-20i}{3+5}

a_{AC}=\left\{2.5i\right\} m/s^2