A long a straight road, a car moving with a speed of 130 km/h is brought to a stop in a distance of 210 m. (a) Find the magnitude of the deceleration of the car ( assumed uniform ). (b) How long does it take for the car to stop ?

Answer

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Finding the Magnitude of Deceleration

1. **Convert the speed from km/h to m/s**:
- The initial speed of the car is given as 130 km/h.
- To convert km/h to m/s, we use the conversion factor:
\[ \text{Speed in m/s} = \text{Speed in km/h} \times \frac{5}{18} \]
- Therefore:
\[ u = 130 \times \frac{5}{18} = 36.1 \, \text{m/s} \]

2. **Identify the known values**:
- Initial velocity \( u = 36.1 \, \text{m/s} \)
- Final velocity \( v = 0 \, \text{m/s} \) (since the car comes to a stop)
- Distance \( s = 210 \, \text{m} \)

3. **Use the equation of motion**:
- We will use the equation:
\[ v^2 = u^2 + 2as \]
- Rearranging the equation to solve for acceleration \( a \):
\[ a = \frac{v^2 - u^2}{2s} \]
- Substituting the known values:
\[ a = \frac{0^2 - (36.1)^2}{2 \times 210} \]
\[ a = \frac{-1303.21}{420} = -3.10 \, \text{m/s}^2 \]

4. **Determine the magnitude of deceleration**:
- The magnitude of deceleration is the absolute value of acceleration:
\[ \text{Magnitude of deceleration} = |a| = 3.10 \, \text{m/s}^2 \]

### Part (b): Finding the Time Taken to Stop

1. **Use another equation of motion**:
- We will use the equation:
\[ v = u + at \]
- Rearranging to solve for time \( t \):
\[ t = \frac{v - u}{a} \]
- Substituting the known values:
\[ t = \frac{0 - 36.1}{-3.10} \]
\[ t = \frac{-36.1}{-3.10} = 11.64 \, \text{s} \]

### Final Answers:
- (a) The magnitude of the deceleration is \( 3.10 \, \text{m/s}^2 \).
- (b) The time taken for the car to stop is \( 11.64 \, \text{s} \).