The Second Law of Reflection states that the angle of incidence is always equal to the angle of reflection. Therefore, the angle of reflection is same as the angle of incidence, that is, 00. So, the reflected ray will retrace the path of the incident ray.

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3 years ago

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates

olasank [31]

2) 20.2 m/s

In the first 4.4 seconds of its motion, the blue car accelerates at a rate of

$a=4.6 m/s^2$

So its final velocity after these 4.4 seconds is

$v=u+at$

where

u = 0 is the initial velocity (the car starts from rest)

a is the acceleration

t is the time

Substituting t = 4.4 s, we find

$v=0+(4.6)(4.4)=20.2 m/s$

After this, the car continues at a constant speed for another 8.5 s, so it will keep this velocity until

$t=4.4 + 8.5 = 12.9 s$

Therefore, the velocity of the car 10.4 seconds after it starts will still be 20.2 m/s.

3) 216.2 m

The distance travelled by the car during the first 4.4 s of the motion is given by

$d_1 = ut_1 + \frac{1}{2}at_1^2$

where

u = 0 is the initial velocity

$t_1 = 4.4 s$ is the time

$a=4.6 m/s^2$ is the acceleration

Substituting,

$d_1 = 0 +\frac{1}{2}(4.6)(4.4)^2=44.5 m$

The car then continues for another 8.5 s at a constant speed, so the distance travelled in the second part is

$d_2 = vt_2$

where

$v_2 = 20.2 m/s$ is the new velocity

$t_2 = 8.5 s$ is the time

Substituting,

$d_2 = (20.2)(8.5)=171.7 m$

So the total distance travelled before the brakes are applied is

$d=44.5 m+171.7 m=216.2 m$

4) $-6.62 m/s^2$

We are told that the blue car comes to a spot at a distance of 247 meters from the start. Therefore, the distance travelled by the car while the brakes are applied is

$d_3 = 247 m -216.2 m=30.8 m$