Question

1. average velocity and average acceleration a canoeist paddles upstream at a velocity of 2.0 m/s for 4.0 s and then floats downstream at 4.0 m/s for 4.0 s. a. what is the average velocity of the canoe during the 8.0 - s time interval? b. what is the average acceleration of the canoe during the 8.0 - s time interval?

Explanation:

Step1: Calculate the displacement upstream

Let the upstream velocity $v_1 = - 2.0\ m/s$ (negative because it's in the opposite - direction of the downstream) and time $t_1=4.0\ s$. Using the formula $d = vt$, the displacement upstream $d_1=v_1t_1=-2.0\times4.0=-8.0\ m$.

Step2: Calculate the displacement downstream

Let the downstream velocity $v_2 = 4.0\ m/s$ and time $t_2 = 4.0\ s$. Using the formula $d = vt$, the displacement downstream $d_2=v_2t_2=4.0\times4.0 = 16.0\ m$.

Step3: Calculate the total displacement

The total displacement $d=d_1 + d_2=-8.0 + 16.0=8.0\ m$. The total time $t=t_1 + t_2=4.0+4.0 = 8.0\ s$.

Step4: Calculate the average velocity

The average - velocity formula is $\bar{v}=\frac{d}{t}$. Substituting the values of $d$ and $t$, we get $\bar{v}=\frac{8.0}{8.0}=1.0\ m/s$.

Step5: Calculate the initial and final velocities

The initial velocity $v_i=-2.0\ m/s$ and the final velocity $v_f = 4.0\ m/s$.

Step6: Calculate the average acceleration

The average - acceleration formula is $\bar{a}=\frac{v_f - v_i}{t}$. Substituting $v_i=-2.0\ m/s$, $v_f = 4.0\ m/s$, and $t = 8.0\ s$, we have $\bar{a}=\frac{4.0-(-2.0)}{8.0}=\frac{4.0 + 2.0}{8.0}=\frac{6.0}{8.0}=0.75\ m/s^2$.

Answer:

a. $1.0\ m/s$
b. $0.75\ m/s^2$