Respuesta :
Answer:
(A) Yes, the driver hits the deer with a speed of 14.66 m/s.
(B) 22.21 m/s
Explanation:
Assume:
- u = initial velocity of the driver
- v = final speed of the driver
- a = acceleration of the driver
- t = time interval for which the brakes were applied
- s = displacement of the driver during brakes application
Part (A):
Before the brakes were applied, the driver moves with a constant velocity of 25 m/s for 1.5 s.
Let us find out distance traveled travel by the driver in this time interval.
[tex]x = 25\times 1.5 = 37.5\ m[/tex]
After the brakes were applied, we have
[tex]u = 25\ m/s\\v = 0\ m/s\\a = -10\ m/s^2\\[/tex]
Using the formula for constant acceleration motion, we have
[tex]v^2=u^2+2as\\\Rightarrow s = \dfrac{v^2-u^2}{2a}\\\Rightarrow s = \dfrac{(0)^2-(25)^2}{2\times (-10)}\\\Rightarrow s = 31.25\ m[/tex]
This means the driver moves a distance of 31.25 m to stop from the time he applies brakes.
So, the total distance traveled by the driver in the entire journey = 37.5 m + 31.25 m = 68.75 m.
Since, the distance traveled by the driver (68.75 m) is greater than the distance of the deer from the driver (58.0 m). So, the driver hits the deer before stopping.
For calculating the speed of the driver with which the driver hits the deer, we have to calculate the actual distance to be traveled by the driver from the instant he applies brake so that the deer does not get hit. This distance is given by:
s = 58 m - 37.5 m = 20.5 m
Now, again using the equation for constant acceleration, we have
[tex]v^2=u^2+2as\\\Rightarrow v^2=(25)^2+2(-10)(20.5)\\\Rightarrow v^2=625-410\\\Rightarrow v^2=215\\\Rightarrow v=\pm \sqrt{215}\\\Rightarrow v=\pm 14.66\ m/s\\[/tex]
Since the driver hits the deer, this means the speed must be positive.
[tex]\therefore v = 14.66 m/s[/tex]
Hence, the driver hits the deer at a speed of 14.66 m/s.
Part (B):
Let the maximum velocity of the driver so that it does not hit the deer be u.
Then,
[tex]x = u\times 1.5 = 1.5u[/tex]
After travelling this distance, the driver applies brakes till it just reaches the position of the deer and does not hit her.
This much distance is S (let).
Using the equation of constant acceleration, we have
[tex]v^2=u^2+2aS\\\Rightarrow (0)^2=u^2+2(-10)(58-x)\\\Rightarrow 0=u^2-1160+30u\\\Rightarrow u^2+30u-1160=0\\[/tex]
On solving the above quadratic equation, we have
[tex]u = 22.21\ m/s\,\,\, or u = -52.21\ m/s[/tex]
But, this speed can not be negative.
So, u = 22.21 m/s.
Hence, the driver could have traveled with a maximum speed of 22.21 m/s so that he does not hit the deer.
