Question

a car moves north at a constant speed of 3 m s^(-1) for 20 s and then east at a constant speed of 4 m s^(-1) for 20 s. what is the average speed of the car during this motion?
a. 7.0 m s^(-1)
b. 5.0 m s^(-1)
c. 3.5 m s^(-1)
d. 2.5 m s^(-1)

Explanation:

Step1: Calculate north - ward distance

Use formula $d = vt$. For north - ward motion, $v_1=3\ m/s$ and $t_1 = 20\ s$, so $d_1=v_1t_1=3\times20 = 60\ m$.

Step2: Calculate east - ward distance

For east - ward motion, $v_2 = 4\ m/s$ and $t_2=20\ s$, so $d_2=v_2t_2=4\times20=80\ m$.

Step3: Calculate total distance

$d=d_1 + d_2=60 + 80=140\ m$.

Step4: Calculate total time

$t=t_1 + t_2=20+20 = 40\ s$.

Step5: Calculate average speed

Use formula $v_{avg}=\frac{d}{t}$. So $v_{avg}=\frac{140}{40}=3.5\ m/s$.

Answer:

C. $3.5\ m/s$