Answer to Question #272789 in Physics for Ton

Question

1. A jeep is traveling at 40 m/s when the breaks are applied. It comes to a stop after 20 seconds. Find the following: (a) acceleration of the jeep (b) distance traveled after the breaks are applied.

2.A car starting from rest moves with constant acceleration. After 10 seconds the velocity is now 30 m/s. What is its acceleration?

3. A car accelerates from rest at a constant rate of 3.5 m/s^2. What is the speed of the car 12 seconds later?

4. A bike traveling at 32.5 m/s skids to a stop in 3 seconds. Determine the skidding distance of the bike.

5. A car is traveling at a constant speed of 25 m/s. It begins to accelerate at a rate of 2.7 m/s^2 for 7.1 seconds. (a) How far does it travel during this time? (b) What is the final speed of the car 7.1 seconds later?

6. You are riding your bike downtown at a speed of 15 m/s and you see a red light ahead. You gently brake over a distance of 50 m and come to a complete stop. What is your acceleration during this time?

Answer 1

1) a)

$$a=\dfrac{v-v_0}{t}=\dfrac{0-40\ \dfrac{m}{s}}{20\ s}=-2\ \dfrac{m}{s^2}.$$

b)

$$d=v_0t+\dfrac{1}{2}at^2,$$ $$d=40\ \dfrac{m}{s}\times20\ s+\dfrac{1}{2}\times(-2\ \dfrac{m}{s^2})\times(20\ s)^2=400\ m.$$

2)

$$a=\dfrac{v-v_0}{t}=\dfrac{30\ \dfrac{m}{s}-0}{10\ s}=3\ \dfrac{m}{s^2}.$$

3)

$$v=v_0+at=0+3.5\ \dfrac{m}{s^2}\times12\ s=42\ \dfrac{m}{s^2}.$$

4)

$$d=\dfrac{1}{2}(v_0+v)t,$$ $$d=\dfrac{1}{2}\times(32.5\ \dfrac{m}{s}+0)\times3\ s=48.75\ m.$$

5)

$$d=v_0t+\dfrac{1}{2}at^2,$$ $$d=25\ \dfrac{m}{s}\times7.1\ s+\dfrac{1}{2}\times2.7\ \dfrac{m}{s^2}\times(7.1\ s)^2=245.5\ m,$$ $$v=v_0+at=25\ \dfrac{m}{s}+2.7\ \dfrac{m}{s^2}\times7.1\ s=44.17\ \dfrac{m}{s}.$$

6)

$$v^2=v_0^2+2ad,$$ $$a=-\dfrac{v_0^2}{2d}=-\dfrac{(15\ \dfrac{m}{s})^2}{2\times50\ m}=-2.25\ \dfrac{m}{s^2}.$$