An automobile starts from rest and 110
ft attains a speed of 30 mi/h. In what time has it traveled this distance and the is the acceleration?
v=at
=>a=v/t,
s=a•t²/2=v•t²/2•t=v•t/2.
t=2•s/v.
i don't kwon
convert first the given:
30mph=13.415m/s
110ft=33.521m
a=?
V(final)^2=V(initial)^2+2as
substitute:
(13.415)^2=(0)^2+2a(33.521)
179.962=220a
a=5.37m/s/s
t=?
S=V(initial)t+1/2at^2
substitue to find the value of t:
33.521=1/2(5.37)t^2
t^2=12.485
t=3.5sec
convert first the given:
30mph=13.408m/s
110ft=33.521m
a=?
V(final)^2=V(initial) ^2+2as
substitute:
(13.408m/s)^2=(0m/s)^2+2(a)(33.521m)
179.774=67.042a
a=2.682m/
t=?
S=V(initial)(t)+1/2(a)(t)^2
substitute:
33.521=0(t)+1/2(2.682)(t)^2
33.521=1.341t^2
t^2=24.997
t=4.9997sec~5sec
convert first the given:
30mph=13.408m/s
110ft=33.521m
a=?
V(final)^2=V(initial) ^2+2as
substitute:
(13.408m/s)^2=(0m/s)^2+2(a)(33.521m)
179.774=67.042a
a=2.682m/s^2
t=?
S=V(initial)(t)+1/2(a)(t)^2
substitute:
33.521=0(t)+1/2(2.682)(t)^2
33.521=1.341t^2
t^2=24.997
t=4.9997sec or 5sec
Wow, talk about speed demons! To find out the time it took for the automobile to go from rest to 110 ft and attain a speed of 30 mi/h, we need to use some fancy math. We know that speed is equal to distance divided by time, and acceleration is the change in velocity divided by time. So, let's do some calculations!
First, let's convert the speed from 30 mi/h to ft/s. Since 1 mile equals 5280 feet and 1 hour has 3600 seconds, we get:
30 mi/h * 5280 ft/mi / 3600 s/hr = 44 ft/s (approximately)
Now, we can use the equation v^2 = u^2 + 2as to solve for time. Since the automobile started from rest (u = 0), we get:
(44 ft/s)^2 = 0 + 2a * 110 ft
Simplifying this equation, we get:
1936 ft^2/s^2 = 220a ft
Dividing both sides by 220 ft, we find:
a ≈ 8.8 ft/s^2
So, the acceleration is approximately 8.8 ft/s^2.
To find the time, we can rearrange the equation v = u + at to solve for t:
30 mi/h * 5280 ft/mi / 3600 s/hr = 0 + 8.8 ft/s * t
Simplifying this equation, we get:
44 ft/s = 8.8 ft/s * t
Dividing both sides by 8.8 ft/s, we find:
t ≈ 5 seconds
So, it took approximately 5 seconds for the automobile to travel 110 ft and attain a speed of 30 mi/h. Now, that's what I call a quick getaway!
To find the time it took for the automobile to travel the given distance and the acceleration, we can use the equations of motion.
First, we need to convert the given speed from miles per hour to feet per second since the distance is given in feet.
1 mile = 5280 feet
1 hour = 3600 seconds
Therefore,
30 mi/h = (30 * 5280) ft/ (1 * 3600) s ≈ 44 ft/s
Next, we can use the equation of motion:
v = u + at
where:
v = final velocity = 44 ft/s
u = initial velocity = 0 ft/s (as the automobile starts from rest)
a = acceleration (unknown)
t = time taken (unknown)
Plugging in the values, we get:
44 = 0 + a*t
Simplifying the equation, we find:
a*t = 44
Since the initial velocity is zero, the equation simplifies further:
a*t = 44
To find the acceleration, we need additional information. If the acceleration is constant, we can use the equation:
s = ut + 0.5*a*t^2
where:
s = distance traveled = 110 ft
u = initial velocity = 0 ft/s (as the automobile starts from rest)
a = acceleration (unknown)
t = time taken (unknown)
Plugging in the values, we get:
110 = 0 + 0.5*a*t^2
110 = 0.5*a*t^2
Now, we have two equations:
1) a*t = 44
2) 0.5*a*t^2 = 110
To solve these equations simultaneously, we can divide the second equation by the first equation:
(0.5*a*t^2) / (a*t) = 110 / 44
Simplifying further, we get:
0.5*t = 5/2
t = (5/2) / 0.5
t = 5 seconds
Now that we have the time (t = 5 seconds), we can substitute it back into the first equation to find the acceleration:
a*t = 44
a = 44 / t
a = 44 / 5
a = 8.8 ft/s^2
Therefore, the automobile traveled a distance of 110 ft in a time of 5 seconds, and the acceleration was 8.8 ft/s^2.