Physics, 14.09.2019 03:30 chicalapingpon1938

You're driving down the highway late one night at 25 m/s when a deer steps onto the road 58min front of you. because you are distracted by your phone, your reaction time before stepping on the brakes is 1.50s, and the maximum deceleration of your car is 10m/s^2. (a) do you hit the deer? if not, how much distance is between you and the deer when you stop? if you hit the deer, how fast are you going? ) (b) what is the maximum initial speed you could have so that you do not hit the deer?

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Answer from: mostman077

(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 driverv = final speed of the drivera = acceleration of the drivert = time interval for which the brakes were applieds = 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.

$x = 25\times 1.5 = 37.5\ m$

After the brakes were applied, we have

$u = 25\ m/s\\v = 0\ m/s\\a = -10\ m/s^2\\$

Using the formula for constant acceleration motion, we have

$v^2=u^2+2as\\\Rightarrow s = \dfrac{v^2-u^2}{2a}\\\Rightarrow s = \dfrac{(0)^2-(25)^2}{2\times (-10)}\\\Rightarrow s = 31.25\ m$

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

$v^2=u^2+2as\\\Rightarrow v^2=(25)^2+2(-10)(20.5)\\\Rightarrow v^2=625-410\\\Rightarrow v^2=215\\\Rightarrow v=\pm \sqrt{215}\\\Rightarrow v=\pm 14.66\ m/s\\$

Since the driver hits the deer, this means the speed must be positive.

$\therefore v = 14.66 m/s$

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,

$x = u\times 1.5 = 1.5u$

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

$v^2=u^2+2aS\\\Rightarrow (0)^2=u^2+2(-10)(58-x)\\\Rightarrow 0=u^2-1160+30u\\\Rightarrow u^2+30u-1160=0\\$

On solving the above quadratic equation, we have

$u = 22.21\ m/s\,\,\, or u = -52.21\ m/s$

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.

Answer from: Quest

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Answer from: Quest

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