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alex41 [277]
3 years ago
13

you are driving at 18m/s down Lyndale​ avenue. A car backs out a driveway 25 meters in front of you. You continue at that speed

while your brain processes the danger which takes 0.25 seconds. You then slam on the brakes and your car has an acceleration of -5.4m/s^2. Do you stop before reaching the driveway(and hitting the car)? Show your work. This is a two-part​ problem with constant speed and braking.
Physics
1 answer:
zimovet [89]3 years ago
7 0

Hello!

For this, first let's calculate time of stop:

t = (V - Vi) / a

Replacing:

t = (0 m/s - 18 m/s) / -5,4 m/s^2

Resolving:

t = -18 m/s / -5,4 m/s^2

t = 3,33 s + 0,25 s = 3,58 s

Now lets calculate distance traveled, with formula:

d = Vi*t + (a*t^2)/2

Replacing:

d = 18 m/s * 3,58 s + (-5,4 m/s^2 * (3,58 s)^2) /2

Resolving:

d = 64,44 m + (-34,604 m)

d = 29,83 m

Then, the vehicle will CRASH

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Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion:        Vf = Vi + a×t

Vf stands for final velocity

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a stands for acceleration

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In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

                       Vf = Vi + a×t

                       Vf = 0 + 2.4×3.6

                       Vf = 2.4×3.6

                       Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

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