A car traveling 60 mi/h along a straight road decelerates at
a constant rate of 11 ft/s2.
(a) How long will it take until the speed is 45 mi/h?
(b) How far will the car travel before coming to a stop?
To solve this problem, we can convert all the given measurements to a consistent unit system. Let's convert the speed from miles per hour (mi/h) to feet per second (ft/s).
Given:
Initial speed (v0) = 60 mi/h
Deceleration (a) = -11 ft/s^2 (negative because it's a deceleration)
Target speed (v) = 45 mi/h
(a) How long will it take until the speed is 45 mi/h?
First, let's convert the speeds to feet per second:
v0 = 60 mi/h = (60*5280 ft) / (1 hour * 3600 s) = 88 ft/s
v = 45 mi/h = (45*5280 ft) / (1 hour * 3600 s) = 66 ft/s
To calculate the time it takes for the car to decelerate from 88 ft/s to 66 ft/s, we can use the following equation:
v = v0 + at
where:
v0 is the initial velocity,
a is the acceleration (deceleration in this case),
t is the time,
and v is the final velocity.
Rearranging the equation to solve for time (t):
t = (v - v0) / a
Plugging in the given values:
t = (66 ft/s - 88 ft/s) / (-11 ft/s^2)
Simplifying, we get:
t = (-22 ft/s) / (-11 ft/s^2)
t = 2 seconds
Therefore, it will take 2 seconds until the speed is 45 mi/h.
(b) How far will the car travel before coming to a stop?
To calculate the distance traveled, we can use the equation for distance:
d = v0t + (1/2)at^2
where:
d is the distance traveled,
v0 is the initial velocity,
a is the acceleration (deceleration in this case),
t is the time, and
v is the final velocity (which will be 0 when the car comes to a stop).
In this case, we want to find the distance traveled when the final velocity (v) is 0.
Plugging in the values:
d = (88 ft/s)(2 s) + (1/2)(-11 ft/s^2)(2 s)^2
Simplifying, we get:
d = 176 ft + (-22 ft/s^2)(4 s^2)
d = 176 ft - 88 ft
d = 88 ft
Therefore, the car will travel 88 feet before coming to a stop.
To find the answers to these questions, we need to use the equations of motion for uniformly decelerating objects.
(a) How long will it take until the speed is 45 mi/h?
Step 1: Convert the initial velocity from miles per hour to feet per second.
1 mile = 5280 feet
1 hour = 3600 seconds
Initial velocity = 60 mi/h * 5280 ft/mi / 3600 s/h ≈ 88 ft/s
Step 2: Convert the target velocity from miles per hour to feet per second.
Target velocity = 45 mi/h * 5280 ft/mi / 3600 s/h ≈ 66 ft/s
Step 3: Use the equation of motion to find the time.
The equation of motion for uniformly decelerating objects is:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is 88 ft/s, the final velocity (v) is 66 ft/s, and the acceleration (a) is the deceleration rate given as 11 ft/s^2.
Rearranging the equation and solving for time (t):
t = (v - u) / a
t = (66 ft/s - 88 ft/s) / -11 ft/s^2 ≈ 2 seconds
So it will take approximately 2 seconds for the car to reach a speed of 45 mi/h.
(b) How far will the car travel before coming to a stop?
Step 1: Convert the deceleration rate from feet per second squared to miles per hour squared.
1 mile = 5280 feet
1 hour = 3600 seconds
Deceleration rate = 11 ft/s^2 * (3600 s/h / 5280 ft/mi)^2 ≈ 7.5 mi/h^2
Step 2: Use the equation of motion to find the distance traveled.
The equation of motion for uniformly decelerating objects is:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity
a = acceleration
t = time
In this case, the initial velocity (u) is 88 ft/s, the acceleration (a) is the deceleration rate given as 11 ft/s^2, and we need to find the time (t) when the car comes to a stop (v = 0).
Rearranging the equation and solving for distance (s):
s = ut + (1/2)at^2
When the car comes to a stop (v = 0), the equation becomes:
0 = (88 ft/s)t + (1/2)(-11 ft/s^2)t^2
Simplifying the equation gives:
-11t^2 + 88t = 0
Factoring out t gives:
t(-11t + 88) = 0
So t = 0 (which means the car hasn't traveled yet) or -11t + 88 = 0
Solving for t:
-11t + 88 = 0
11t = 88
t ≈ 8 seconds
Now substitute the value of t in the equation of motion to find the distance:
s = ut + (1/2)at^2
s = (88 ft/s)(8 s) + (1/2)(-11 ft/s^2)(8 s)^2
s = 704 ft + (1/2)(-11 ft/s^2)(64 s^2)
s = 704 ft - 352 ft
s ≈ 352 ft
So the car will travel approximately 352 feet before coming to a stop.
Vi = 60 mi/h * 5280 ft/mi /3600 s/h
Vf = 45 * 5280/3600
Vf = Vi - 11 t
solve for t
===========================
0 = Vi - 11 T
T = Vi/11
d = Vi T - 5.5 T^2