Question

2 motion with constant acceleration (continued)

1. distance an in - line skater first accelerates from 0.0 m/s to 5.0 m/s in 4.5 s, then continues at this constant speed for another 4.5 s. what is the total distance traveled by the in - line skater?
2. final velocity a plane travels a distance of 5.0×10² m north while being accelerated uniformly from rest at the rate of 5.0 m/s². what final velocity does it attain?
3. final velocity an airplane accelerated uniformly from rest at the rate of 5.0 m/s² south for 14 s. what final velocity did it attain?

Explanation:

36.

Step1: Calculate distance during acceleration

Use the formula $d_1 = v_0t+\frac{1}{2}at^2$. First find acceleration $a=\frac{v - v_0}{t}=\frac{5.0 - 0.0}{4.5}=\frac{10}{9}\text{ m/s}^2$. Since $v_0 = 0\text{ m/s}$, then $d_1=\frac{1}{2}\times\frac{10}{9}\times(4.5)^2 = 11.25\text{ m}$.

Step2: Calculate distance during constant - speed

Use the formula $d_2=v\times t$. Here $v = 5.0\text{ m/s}$ and $t = 4.5\text{ s}$, so $d_2=5.0\times4.5 = 22.5\text{ m}$.

Step3: Calculate total distance

$d=d_1 + d_2=11.25+22.5 = 33.75\text{ m}$.

Step1: Use the kinematic equation

Use the equation $v^2=v_0^2 + 2ad$. Given $v_0 = 0\text{ m/s}$, $a = 5.0\text{ m/s}^2$ and $d=5.0\times10^{2}\text{ m}$.

Step2: Solve for final velocity

$v^2=0 + 2\times5.0\times5.0\times10^{2}$, so $v^2 = 5000$. Then $v=\sqrt{5000}\approx70.7\text{ m/s}$.

Step1: Use the kinematic equation

Use the equation $v=v_0+at$. Given $v_0 = 0\text{ m/s}$, $a = 5.0\text{ m/s}^2$ and $t = 14\text{ s}$.

Step2: Solve for final velocity

$v=0+5.0\times14=70\text{ m/s}$.

Answer:

$33.75\text{ m}$

37.