An automobile traveling along a straight road increases its speed from 67 ft/s to 85 ft/s in 180 ft. If the acceleration is constant, how much
time elapses while the auto moves the 180 ft? Answer in units of s. Can you tell me the steps in doing this?
To find the time elapsed while the automobile moves the distance of 180 ft, we can use the equation:
v = u + at
where:
v = final velocity (85 ft/s)
u = initial velocity (67 ft/s)
a = acceleration (which is constant)
t = time elapsed
First, we need to find the acceleration using the given information. The change in velocity (∆v) is 85 ft/s - 67 ft/s = 18 ft/s. The distance (∆s) covered is 180 ft.
We can use the kinematic equation:
∆v = u + at
Substituting the values, we have:
18 ft/s = 67 ft/s + a * t
Rearranging the equation, we get:
18 ft/s - 67 ft/s = a * t
-49 ft/s = a * t
Next, we can calculate the acceleration (a) using the known distance (∆s) and the equation:
∆s = u * t + 1/2 * a * t^2
Since the initial velocity (u) is 67 ft/s and the distance (∆s) is 180 ft, we can solve for a:
180 ft = 67 ft/s * t + 1/2 * a * t^2
Now, we substitute this value of a into the previous equation:
-49 ft/s = (180 ft - 67 ft/s * t) * t
Simplifying the equation gives a quadratic equation:
49 t^2 - 67 t - 180 = 0
We can solve this equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For the quadratic equation 49 t^2 - 67 t - 180 = 0, we have:
a = 49, b = -67, c = -180
Substituting these values into the quadratic formula:
t = (-(-67) ± √((-67)^2 - 4 * 49 * -180)) / (2 * 49)
Simplifying further:
t = (67 ± √(4489 + 35280)) / 98
t = (67 ± √39769) / 98
Taking the positive value of the square root:
t = (67 + √39769) / 98
Now we can calculate this value:
t = (67 + 199) / 98
t = 266 / 98
t ≈ 2.714 s
Therefore, the time elapsed while the automobile moves the 180 ft is approximately 2.714 seconds.
vf^2=vi^2 + 2 a d
solve for a
then
d=vi*t+1/2 a t^2 solve for time with d=180ft