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You are driving your car along a country road at a speed of 27.0 m/s. as you come over the crest of a hill, you notice a farm tractor 25.0 m ahead of you on the road, moving in the same direction as you at a speed of 10.0 m/s. you immediately slam on your brakes and slow down with a constant acceleration of magnitude 7.00 m/s2 .
a.will you hit the tractor before you stop?
b.how far will you travel before you stop or collide with the tractor?
c.if you stop, how far is the tractor in front of you when you finally stop?

Respuesta :

speed of the car = 27 m/s

speed of truck ahead = 10 m/s

relative speed of car with respect to truck

[tex]v_r = 27 - 10 = 17 m/s[/tex]

relative deceleration of car

[tex]a_r = -7 m/s^2[/tex]

now the distance before they stop with respect to each other is given by

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 17^2 = 2 *(-7)*d[/tex]

[tex]d = 20.6 m[/tex]

so it will come at the same speed of truck after 20.6 m distance and hence it will not hit the truck as the distance of the truck is 25 m from car

Part b)

Distance traveled by car before it stops is given by

[tex]v_f^2 - v_i^2 = 2 a s[/tex]

[tex]0^2  - 27^2 = 2 * (-7)* s[/tex]

[tex]s = 52.1 m[/tex]

so it will stop after it will cover total 52.1 m distance

Part c)

time taken by the car to stop

[tex]v_f - v_i = at[/tex]

[tex]0 - 27 = (-7) * t[/tex]

[tex]t = 3.86 s[/tex]

now the distance covered by truck in same time

[tex]d = 3.86 * 10 = 38.6 m[/tex]

now after the car will stop its distance from the truck is

[tex]D = 25 + 38.6 - 52.1 = 11.5 m[/tex]

so the distance between them is 11.5 m

a. The car will not hit the tractor.

b. The car would travel a distance of 52.07 m before stopping

c. At the moment when the car stopped, the tractor is 11.5 m in front of the car.

Linear motion

From the question, we are to determine of you will hit the tractor before you stop

First, we will determine the time it will take the car to stop

From the given information,

Initial velocity, u = 27.0 m/s

a = -7 m/s² (Negative sign indicates deceleration)

v = 0 m/s (Since the car will come to stop)

From one of the equations of linear motion,

[tex]v = u +at[/tex]

Where v is the final velocity

u is the initial velocity

a is the acceleration

and t is the time taken

Putting the parameters into the equation, we get

[tex]0 = 27.0 + (-7)t[/tex]

[tex]7t = 27[/tex]

[tex]t = \frac{27}{7}[/tex]

t = 3.857 secs

This is the time it will take the car to stop

Now, we will determine the distance the car would travel after applying the brakes

Using the formula

[tex]S = \frac{u+v}{2} \times t[/tex]

Where S is the distance traveled

[tex]S = \frac{27.0 + 0}{2} \times 3.857[/tex]

[tex]S = 13.5 \times 3.857[/tex]

S = 52.0695 m

S ≅ 52.07 m

This means the car would travel 52.07 m after applying the brakes

Now, we will determine the distance the tractor would have traveled when the car came to a stop

Speed of the tractor = 10.0 m/s

Time taken for the car to stop = 3.857 secs

Using the formula,

Distance = Speed × Time

Distance = 10.0 × 3.857

Distance = 38.57 m

Now, we will determine the distance between the car and the tractor when the car finally stopped

Distance between the car and the tractor = Distance ahead + Distance traveled by the tractor after the car stopped - Distance traveled by car after applying the brakes

∴ Distance between the car and the tractor = 25 m + 38.57 m - 52.07 m

Distance between the car and the tractor = 11.5 m

Hence,

a. Since the distance the tractor was ahead of the car + the distance the tractor traveled after the car stopped is more than the distance the car traveled after applying the brakes (25 m + 38.57 m > 52.07), the car will not hit the tractor.

b. The car would travel a distance of 52.07 m before stopping

c. At the moment when the car stopped, the tractor is 11.5 m in front of the car.

Learn more on Linear motion here: https://brainly.com/question/6202541

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