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.
Image from: R. C. Hibbeler, K. B. Yap, and S. C. Fan, Mechanics for Engineers: Dynamics (SI Edition), 13th ed. Singapore: Pearson Education South Asia, 2013.
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