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?
-when is the acceleration of the object positive and negative?
v=at
=>a=v/t,
s=at²/2=v t²/2t=vt/2.
t=2s/v.
positive
To find the time it took to travel a certain distance and the acceleration, we can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (converted to ft/s)
u = initial velocity (converted to ft/s)
a = acceleration (ft/s^2)
s = distance (ft)
First, let's convert the speed from miles per hour to feet per second:
30 mi/h = 30 * 5280 ft / 3600 s = 44 ft/s
Since the car starts from rest, the initial velocity (u) is 0 ft/s.
Plugging in the values into the equation:
(44 ft/s)^2 = (0 ft/s)^2 + 2a * 110 ft
1936 ft^2/s^2 = 220a ft
Now we can solve for the acceleration (a):
a = 1936 ft^2/s^2 / 220 ft ≈ 8.8 ft/s^2
Next, let's find the time it took to travel the given distance:
v = u + at
Since the initial velocity is 0 ft/s, the equation simplifies to:
v = at
t = v / a
t = 110 ft / 44 ft/s ≈ 2.5 s
Therefore, it took approximately 2.5 seconds for the automobile to travel the distance of 110 ft, and the acceleration is approximately 8.8 ft/s^2.
Regarding when the acceleration of the object is positive and negative, in this case, the acceleration is constant and positive throughout the motion. The car starts from rest and increases its velocity, so the acceleration remains positive from the start until it reaches the final velocity.