A man driving his car into 20.0 ft garage with a velocity of 20.0mi/h applies the brakes, producing a constant deceleration. Find the smallest deceleration necessary to avoid striking the back wall of the garage. And find how many seconds it takes for the car to come to rest
Vo=20mi/h * 5280Ft/mi*1h/3600s=29.33Ft/s
a = (V^2-Vo^2)/2d
a = (0-29.33^2)/40 = -21.51 Ft/s^2.
V = Vo + a*t = 0
29.33 - 21.51*t = 0
21.51t = 29.33
t = 1.36 s. To stop.
3s
To find the minimum deceleration necessary to avoid hitting the back wall of the garage, we can use the following equations of motion:
1. The equation for displacement with constant acceleration:
s = ut + (1/2)at^2
2. The equation for final velocity with constant acceleration:
v = u + at
Given:
Initial velocity, u = 20.0 mi/h
Final velocity, v = 0 mi/h (since the car comes to rest)
Displacement, s = 20.0 ft
Acceleration, a = ?
First, let's convert the given values to consistent units. We'll use the conversion:
1 mi/h = 1.46667 ft/s
u = 20.0 mi/h × 1.46667 ft/s/mi ≈ 29.333 ft/s
v = 0 mi/h × 1.46667 ft/s/mi = 0 ft/s
Now we can use the second equation to solve for acceleration:
0 = 29.333 ft/s + a × t
a × t = -29.333 ft/s
Since the car comes to rest, the final velocity is 0. Solving the equation, we get:
t = -29.333 ft/s / a (Equation 1)
Next, use the first equation to find the displacement:
s = (u × t) + (1/2) a t^2
20.0 ft = (29.333 ft/s × t) + (1/2) a t^2
Substitute the value of t from Equation 1 into this equation:
20.0 ft = (29.333 ft/s × (-29.333 ft/s / a)) + (1/2) a (-29.333 ft/s / a)^2
Simplifying further:
20.0 ft = (-29.333^2 ft^2/s^2 / a) - (29.333^2 ft^3/s^2 / a^2)
Multiply through by a, and rearrange the equation to solve for a:
a^2 - (29.333^2 ft^2/s^2 / 20.0 ft) a - 29.333^2 ft^2/s^2 = 0
Now we have a quadratic equation in terms of a. We can use the quadratic formula to solve for a:
a = [-(-29.333^2 ft^2/s^2 / 20.0 ft) ± √((-29.333^2 ft^2/s^2 / 20.0 ft)^2 - 4(1)(-29.333^2 ft^2/s^2))] / (2)
Evaluating this expression will yield two possible values for a, but we are only interested in the positive value since deceleration is always negative (opposite direction to velocity). The positive value corresponds to the required minimum deceleration.
To find the time it takes for the car to come to rest, we can substitute this positive value of a into Equation 1 and solve for t.
By following these steps and performing the necessary calculations, you can find the smallest deceleration necessary to avoid hitting the back wall of the garage and determine the time it takes for the car to come to rest.