Question

On a foggy day, two drivers spot each other when they are 80m apart. they are travelling at 72km/h and 60km/h. both of them simultaneously apply brakes which retards both the cars at a rate of 5m/s^2. determine whether they avert collision or not?

Answer 1

Answer on Question #42915 – Physics - Mechanics | Kinematics | Dynamics

On a foggy day, two drivers spot each other when they are 80m apart. they are travelling at 72km/h and 60km/h. both of them simultaneously apply brakes which retards both the cars at a rate of 5m/s^2. determine whether they avert collision or not?

Solution:

$v_{1} = 72\frac{\mathrm{km}}{\mathrm{h}} = 20\frac{\mathrm{m}}{\mathrm{s}}$ – velocity of the first car;

$v_{2} = 60\frac{\mathrm{km}}{\mathrm{h}} = 16.7\frac{\mathrm{m}}{\mathrm{s}}$ – velocity of the second car;

$a = 5\frac{\mathrm{m}}{\mathrm{s}^2}$ – deceleration of the cars;

$D = 80\mathrm{m}$ – initial distance between two cars.

Drivers can avert collision if the sum of the braking distances will be less than 80m. Lets find the distance travelled by both before they come to a halting stop.

Rate equation for the first car (final velocity is zero, $t_1$ – time of breaking):

$$0 = v_1 - a t_1$$ $$t_1 = \frac{v_1}{a}$$

Equation of motion for the first car ($S_1$ – braking distance for the first car):

$$S_1 = v_1 t_1 - \frac{a t_1^2}{2} = |\text{using (1)}| = v_1 \frac{v_1}{a} - \frac{a}{2} \left(\frac{v_1}{a}\right)^2 = \frac{v_1^2}{2a}$$

Rate equation for the second car (final velocity is zero, $t_2$ – time of breaking):

$$0 = v_2 - a t_2$$ $$t_2 = \frac{v_2}{a}$$

Equation of motion for the first car ($S_2$ – braking distance for the second car):

$$S_2 = v_2 t_2 - \frac{a t_2^2}{2} = |\text{using (3)}| = v_2 \frac{v_2}{a} - \frac{a}{2} \left(\frac{v_2}{a}\right)^2 = \frac{v_2^2}{2a}$$ $$S = S_1 + S_2$$

(2) and (4) in (5):

$$S = \frac{v_1^2}{2a} + \frac{v_2^2}{2a} = \frac{v_1^2 + v_2^2}{2a} = \frac{\left(16.7\frac{\mathrm{m}}{\mathrm{s}}\right)^2 + \left(20\frac{\mathrm{m}}{\mathrm{s}}\right)^2}{2 \cdot 5\frac{\mathrm{m}}{\mathrm{s}^2}} = 67.9\ \mathrm{m}$$ $$S < D \Rightarrow$$

Since they are 80 m apart and sum of both their distance is less than 80 m, they would stop before time. There would no collision.

Answer: drivers will avert the collision.

http://www.AssignmentExpert.com/