To solve this problem, we can use the equations of motion, particularly the one that relates initial velocity, acceleration, time, and distance.
- Distance covered in the first 5 seconds:
Given:
- Initial velocity (\(u\)) = \(10, \text{m/s}\)
- Acceleration (\(a\)) = \(-1.2, \text{m/s}^2\) (negative because the brakes are applied and it's deceleration)
- Time (\(t\)) = \(5, \text{s}\)
We use the equation: \[ s = ut + \frac{1}{2}at^2 \]
Substituting the given values into the equation: \[ s = (10, \text{m/s})(5, \text{s}) + \frac{1}{2}(-1.2, \text{m/s}^2)(5, \text{s})^2 \] \[ s = 50, \text{m} + \frac{1}{2}(-1.2)(25) \] \[ s = 50, \text{m} - 15, \text{m} \] \[ s = 35, \text{m} \]
So, the distance the car covers in the first 5 seconds after the acceleration begins is 35 meters.
- Distance the car travels before it comes to a stop:
To find the total distance the car travels before it comes to stop, first we need to find the time it takes to stop. The final velocity (\(v\)) when it stops is \(0, \text{m/s}\).
We can use the formula: \[ v = u + at \] Setting \(v = 0\): \[ 0 = 10 + (-1.2)t \] \[ 1.2t = 10 \] \[ t = \frac{10}{1.2} \approx 8.33, \text{s} \]
Now, we can calculate the total distance traveled before stopping using the formula: \[ s = ut + \frac{1}{2}at^2 \]
Substituting \(u=10, \text{m/s}\), \(a=-1.2, \text{m/s}^2\), and \(t \approx 8.33, \text{s}\): \[ s = (10)(8.33) + \frac{1}{2}(-1.2)(8.33)^2 \] Calculating \(10 \times 8.33\): \[ s = 83.3 + \frac{1}{2}(-1.2)(69.44) \] Calculating \(\frac{1}{2}(-1.2)(69.44)\): \[ s = 83.3 - 41.67 \] \[ s \approx 41.63, \text{m} \]
Thus, the total distance the car travels before it comes to a stop is approximately 41.63 meters.