Question

1. Draw a straight line.

2. Place point a at the beginning of the line.
3. Place point b on the line after a distance of 10 from a.
4. Place point c after a distance of 10 from b
5. Place point d after a distance of 9.5 from c.
6. Place point e after a distance of 10 from d.
7. Place point f after a distance of 2 from e.
Now imagine that this line is a road and a car parking before point a begins to move at a constant speed. Once it reaches point a, its speed begins to deaccelerate according to a function.
Assume that the time required to travel from a to b was X.
The car passes point b and continues deaccelerating according to the function. However, the time to travel from b to c is unspecified.
The car continues past point c and eventually reaches d in X time.
The travel time from d to e is unspecified.
The travel time from e to f is still X.

Assuming that the car never accelerates and continues to deaccelerate in the distances where travel time is not specified: what is the deacceleration function (if possible) such that the car could maintain the travel time of (X) across the specified distances between a and b, c and d, and e and f?
a. (f(t) = -kt)
b. (f(t) = k/t)
c. (f(t) = e⁻ᵏᵗ)
d. (f(t) = √kt)

Answer 1

The deceleration function that allows the car to maintain a constant travel time across the specified distances is f(t) = k/t (Option b)

Step-by-step explanation:

The car's motion can be described using kinematic equations for constant acceleration. Since the car starts from rest and decelerates, we can use the equation:

v = u + at

Where v is the final velocity, u is the initial velocity (0 m/s), and a is the constant deceleration. Using the given information that the time to travel from a to b is X, we can find the deceleration:

v = Xa

Plugging in the values for point a and b (u = 0, v = 10 m, t = X), we get:

10 = Xa

Solving for a, we find:

a = 10 / X

Therefore, the deceleration function is f(t) = k/t (Option b).