Question

A driver,drives a truck of mass 10 ton at a velocity of 72km/hr sees an obstruction on the road at a distance of 145m.if person at reaction of driver is in 0.8second,find weather accident will took place or not.if the retarding force of brake is 16000n.

Answer 1

QUESTION:

A driver, drives a truck of mass 10 ton at a velocity of $72\mathrm{km/hr}$ sees an obstruction on the road at a distance of $145\mathrm{m}$. If person at reaction of driver is in 0.8 second, find weather accident will took place or not. If the retarding force of brake is 16000n.

SOLUTION:

During the $t_r = 0.8$ second (the time of reaction) the truck continues to move at constant velocity $\upsilon_0$ and travels the distance:

$$s_1 = \upsilon_0 t_r$$

Then, when a driver begins to brake, the truck begins to move with deceleration. We find the value of deceleration using Newton's second law:

$$F = m \cdot a$$ $$a = \frac{F}{m}$$

Here $F$ is the force of brake and $m$ is the truck's mass.

So, now we can find the time $t_s$, it takes the truck to stop since driver begins to brake. The truck's velocity $\upsilon_t$ depends on time according to the following equation (because it is motion with constant acceleration):

$$\upsilon_t = \upsilon_0 - a \cdot t$$

And when $t = t_s$ truck's velocity is $v_t = 0$, therefore

$$0 = \upsilon_0 - a \cdot t_s$$ $$t_s = \frac{\upsilon_0}{a} = \frac{m \upsilon_0}{F}$$

The distance traveled since driver begins to brake depends on time according to the following equation (because it is motion with constant acceleration):

$$s_2 = \upsilon_0 t - \frac{a \cdot t^2}{2}$$

Substituting $t$ with $t_s$ we find the distance, the truck travels since driver begins to brake and until truck stops:

$$s_2 = \upsilon_0 \cdot \frac{m \upsilon_0}{F} - \frac{F}{2m} \cdot \frac{m^2 \upsilon_0^2}{F^2} = \frac{m \upsilon_0^2}{F} - \frac{m \upsilon_0^2}{2F} = \frac{m \upsilon_0^2}{2F}$$

(There is another way to obtain the previous equation – without finding the time $t_s$ :

$\upsilon^2 = \upsilon_0^2 + 2 \cdot a \cdot s_2$ , since the final velocity of the truck is equal to zero, and $a < 0$ and $|a| = \frac{F}{m}$ we get $s_2 = \frac{\upsilon_0^2}{2a} = \frac{\upsilon_0^2}{2\frac{F}{m}} = \frac{m \upsilon_0^2}{2F}$ .

Total distance traveled

$$s _ {\text {t o t}} = s _ {1} + s _ {2} = v _ {0} t _ {r} + \frac {m v _ {0} ^ {2}}{2 F}$$

(72 km/hr=20 m/s)

$$s _ {\text {t o t}} = 2 0 \cdot 0. 8 + \frac {1 0 \cdot 1 0 ^ {3} \cdot 2 0 ^ {2}}{2 \cdot 1 6 \cdot 1 0 ^ {3}} = 1 4 1 \mathrm {m}.$$

Hence, the truck will stop before an obstruction.

**ANSWER**

The accident won't took place