A car travelling at `36 km//h` speeds upto `72 km//h` in `5` seconds. What is its acceleration ? If the same car stops in `20` seconds , what is the acceleration?

Explanation

To solve the given problem, we need to calculate the acceleration of the car in two different scenarios. Let's break down the problem step by step.

### Part 1: Acceleration when the car speeds up

1. **Given Data:**
- Initial speed (\(u\)) = 36 km/h
- Final speed (\(v\)) = 72 km/h
- Time (\(t\)) = 5 seconds

2. **Convert speeds from km/h to m/s:**
- \(u = 36 \, \text{km/h} \times \frac{5}{18} = 10 \, \text{m/s}\)
- \(v = 72 \, \text{km/h} \times \frac{5}{18} = 20 \, \text{m/s}\)

3. **Use the first equation of motion to find acceleration (\(a\)):**
\[ v = u + at \]
Substitute the known values:
\[ 20 = 10 + a \times 5 \]

4. **Solve for \(a\):**
\[ 20 - 10 = 5a \]
\[ 10 = 5a \]
\[ a = \frac{10}{5} = 2 \, \text{m/s}^2 \]

So, the acceleration of the car when it speeds up is \(2 \, \text{m/s}^2\).

### Part 2: Acceleration when the car stops

1. **Given Data:**
- Initial speed (\(u\)) = 72 km/h (which is 20 m/s as calculated earlier)
- Final speed (\(v\)) = 0 m/s (since the car stops)
- Time (\(t\)) = 20 seconds

2. **Use the first equation of motion to find acceleration (\(a'\)):**
\[ v = u + a't \]
Substitute the known values:
\[ 0 = 20 + a' \times 20 \]

3. **Solve for \(a'\):**
\[ 0 = 20 + 20a' \]
\[ -20 = 20a' \]
\[ a' = \frac{-20}{20} = -1 \, \text{m/s}^2 \]

So, the acceleration of the car when it stops is \(-1 \, \text{m/s}^2\). The negative sign indicates deceleration.

### Summary:
- Acceleration when speeding up: \(2 \, \text{m/s}^2\)
- Acceleration when stopping: \(-1 \, \text{m/s}^2\)