Question

The driver of a car travelling at a constant speed of 72 kmph, seeing a procession ahead of him moving in the same direction, decelerates at a constant rate of 1m/s^2 until his speed is reduced to 3 kmph. He moves at this speed for 2 minutes until the procession takes a left turn. He then accelerates at a constant rate of 0.5 m/s^2 to his normal speed. Determine the total distance travelled during this time and the time lost.

Answer 1

Answer on Question #71713, Physics / Other

The driver of a car travelling at a constant speed of $72\,\mathrm{kmph}$, seeing a procession ahead of him moving in the same direction, decelerates at a constant rate of $1\,\mathrm{m/s^2}$ until his speed is reduced to $3\,\mathrm{kmph}$. He moves at this speed for 2 minutes until the procession takes a left turn. He then accelerates at a constant rate of $0.5\,\mathrm{m/s^2}$ to his normal speed. Determine the total distance travelled during this time and the time lost.

Solution

$72\,\mathrm{km/h} = 20\,\mathrm{m/s}$

$3\,\mathrm{km/h} = 0.83\,\mathrm{m/s}$

**Deceleration stage**

$$t_1 = \frac{0.83 - 20}{-1} = 19.17\,\mathrm{s};$$ $$d_1 = \frac{0.83^2 - 20^2}{2 \times (-1)} = 199.66\,\mathrm{m}$$

**Motion at constant speed**

$$t_2 = 120\,\mathrm{s};$$ $$d_2 = 120 \times 0.83 = 99.6\,\mathrm{m}$$

**Acceleration stage**

$$t_3 = \frac{20 - 0.83}{0.5} = 38.34\,\mathrm{s};$$ $$d_1 = \frac{20^2 - 0.83^2}{2 \times 0.5} = 399.31\,\mathrm{m}$$

**Total distance**

$$d = 199.66 + 99.6 + 399.31 = 698\,\mathrm{m}$$

**Time to move same distance at normal speed**

$$t = \frac{698}{20} = 34.9\,\mathrm{s}$$

**Time lost**

$$\Delta t = 19.17 + 120 + 38.34 - 34.9 = 142.61\,\mathrm{s} = 2.4\,\mathrm{min}$$

Answer: 698 m; 2.4 min.

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