Question

1. what is the average velocity of a bike that travels 210 m e and then travels 300 m w in 30 s? 3 marks t1
2. a driver is travelling at 25 m/s forward when she spots a sign that reads \bridge out ahead.\ after applying the brakes, the car slows down at a rate of 3.0 m/s².

(a) how much time was required to stop the car once the brakes were applied? 4 marks
(b) how far did the car travel after the brakes were applied? 4 marks t1

Explanation:

11.

Step1: Define displacement

Displacement is the net change in position. The bike moves 210 m east and 300 m west. Taking east as positive and west as negative, the net displacement $d=210 - 300=- 90$ m.

Step2: Use average - velocity formula

The formula for average velocity is $v_{avg}=\frac{d}{t}$, where $d$ is displacement and $t$ is time. Given $t = 30$s. Then $v_{avg}=\frac{-90}{30}=-3$ m/s. The negative sign indicates the direction is west.

Step1: Identify the kinematic - equation

We use the kinematic equation $v = v_0+at$, where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration and $t$ is the time. The car stops, so $v = 0$, $v_0=25$ m/s and $a=-3.0$ m/s² (negative because it is decelerating).

Step2: Solve for time

Rearranging the equation $v = v_0+at$ for $t$ gives $t=\frac{v - v_0}{a}$. Substituting the values: $t=\frac{0 - 25}{-3}=\frac{25}{3}\approx8.33$ s.

Step1: Identify the kinematic - equation

We use the kinematic equation $v^{2}=v_{0}^{2}+2ad$, where $v$ is the final velocity, $v_0$ is the initial velocity, $a$ is the acceleration and $d$ is the displacement. Here, $v = 0$, $v_0 = 25$ m/s and $a=-3.0$ m/s².

Step2: Solve for displacement

Rearranging the equation $v^{2}=v_{0}^{2}+2ad$ for $d$ gives $d=\frac{v^{2}-v_{0}^{2}}{2a}$. Substituting the values: $d=\frac{0 - 25^{2}}{2\times(-3)}=\frac{- 625}{-6}\approx104.17$ m.

Answer:

The average velocity is 3 m/s [W].

12. (a)