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
