Question

quiz on graphs of kinematics physics 1. a ball is placed at rest at the top of a hill. it travels with constant acceleration for the first 12 seconds and reaches a speed of 4 m/s. it then decelerates at a constant rate of 0.1m/s² for 20 seconds. it then travels at a constant speed for a further 18 seconds.

Explanation:

Since the problem seems incomplete (maybe asking to draw a speed - time graph, find total distance, or other kinematic quantities), let's assume we need to find the final speed after deceleration and then the total distance covered.

Step 1: Find the acceleration during the first phase

The ball starts from rest (\(u = 0\ m/s\)), final speed \(v=4\ m/s\) and time \(t = 12\ s\). Using the formula \(a=\frac{v - u}{t}\)
\(a=\frac{4 - 0}{12}=\frac{1}{3}\ m/s^{2}\)

Step 2: Find the speed after deceleration

We know the initial speed for deceleration \(u = 4\ m/s\), deceleration \(a=- 0.1\ m/s^{2}\) (negative because it's deceleration) and time \(t = 20\ s\). Using the formula \(v=u+at\)
\(v = 4+( - 0.1)\times20=4 - 2=2\ m/s\)

Step 3: Calculate the distance covered in each phase

• Phase 1 (Acceleration): Using the formula \(s_1=\frac{u + v}{2}\times t\) (since it's uniformly accelerated motion), \(u = 0\), \(v = 4\), \(t = 12\)

\(s_1=\frac{0 + 4}{2}\times12=2\times12 = 24\ m\)

• Phase 2 (Deceleration): Using the formula \(s_2=ut+\frac{1}{2}at^{2}\), \(u = 4\), \(a=-0.1\), \(t = 20\)

\(s_2=4\times20+\frac{1}{2}\times(- 0.1)\times20^{2}=80-20 = 60\ m\)

• Phase 3 (Constant speed): The speed here is the speed after deceleration \(v = 2\ m/s\) and time \(t = 18\ s\). Using \(s_3=v\times t\)

\(s_3=2\times18 = 36\ m\)

Step 4: Calculate total distance

Total distance \(s=s_1 + s_2+s_3\)
\(s=24 + 60+36=120\ m\)

If the question was about the final speed after deceleration, the answer is \(2\ m/s\). If it was about total distance, the answer is \(120\ m\). Since the original problem is incomplete, please provide the complete question (e.g., draw the speed - time graph, find total distance, find final speed etc.) to get a more accurate answer.

Answer:

Since the problem seems incomplete (maybe asking to draw a speed - time graph, find total distance, or other kinematic quantities), let's assume we need to find the final speed after deceleration and then the total distance covered.

Step 1: Find the acceleration during the first phase

The ball starts from rest (\(u = 0\ m/s\)), final speed \(v=4\ m/s\) and time \(t = 12\ s\). Using the formula \(a=\frac{v - u}{t}\)
\(a=\frac{4 - 0}{12}=\frac{1}{3}\ m/s^{2}\)

Step 2: Find the speed after deceleration

We know the initial speed for deceleration \(u = 4\ m/s\), deceleration \(a=- 0.1\ m/s^{2}\) (negative because it's deceleration) and time \(t = 20\ s\). Using the formula \(v=u+at\)
\(v = 4+( - 0.1)\times20=4 - 2=2\ m/s\)

Step 3: Calculate the distance covered in each phase

• Phase 1 (Acceleration): Using the formula \(s_1=\frac{u + v}{2}\times t\) (since it's uniformly accelerated motion), \(u = 0\), \(v = 4\), \(t = 12\)

\(s_1=\frac{0 + 4}{2}\times12=2\times12 = 24\ m\)

• Phase 2 (Deceleration): Using the formula \(s_2=ut+\frac{1}{2}at^{2}\), \(u = 4\), \(a=-0.1\), \(t = 20\)

\(s_2=4\times20+\frac{1}{2}\times(- 0.1)\times20^{2}=80-20 = 60\ m\)

• Phase 3 (Constant speed): The speed here is the speed after deceleration \(v = 2\ m/s\) and time \(t = 18\ s\). Using \(s_3=v\times t\)

\(s_3=2\times18 = 36\ m\)

Step 4: Calculate total distance

Total distance \(s=s_1 + s_2+s_3\)
\(s=24 + 60+36=120\ m\)

If the question was about the final speed after deceleration, the answer is \(2\ m/s\). If it was about total distance, the answer is \(120\ m\). Since the original problem is incomplete, please provide the complete question (e.g., draw the speed - time graph, find total distance, find final speed etc.) to get a more accurate answer.