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A car is traveling at 50mi/hr when the brakes are fully applied, producing a constant deceleration of 22ft/sec^2. What is the distance covered before the car comes to a stop? Thanks.
How would you use the anitderivative to solve this problem?
A car is traveling at 50mi/hr when the brakes are fully applied, producing a constant deceleration of 22ft/sec^2. What is the distance covered before the car comes to a stop? Thanks.
The general equation which describes this motion is:Originally Posted by UMStudent
s(t) = -(1/2)at^2+v_0(t)+s_0
Where,
s(t) is distance traveled from starting point.
"a" is the acceleration.
v_0 is initial velocity.
s_0 is initial starting distance traveled (which is zero here).
Thus,
s(t) = - (1/2)*11*t^2+50*t
s(t) = -11*t^2+50t
Use this to solve all your problems.
I am going to use a coordinate system where the origin is at the point where deceleration started and +x is the direction of the initial velocity.
a = -22 ft/s^2 and Initial speed = 50 mi/hr = 73.33 ft/s
(a is negative since it is opposing the direction of the initial velocity.)
Now:
v(t) = Int[a(t'), 0, t] + v0 <-- Where v0 is an integration constant.
v(t) = Int[-22, 0, t] + v0 = -22t + v0
Now, at t = 0 s, v(0) = 73.33 ft/s, so
v(t) = -22t + 73.33 ft/s
Again:
x(t) = Int[v(t'), 0, t] + x0 <-- Where x0 is an integration constant.
x(t) = Int[-22t' + 73.33, 0, t]
x(t) = (1/2)(-22)t^2 + 73.33t + x0
At t = 0 s, x(0) = 0 ft, so
x(t) = -11t^2 + 73.33t
When does the car come to a stop? When v(t) = 0 ft/s. Thus when:
v(t) = -22t + 73.33 = 0
t = 73.33/22 s = 3.333 s
So where is the car at this time?
x(3.333) = -11(3.333)^2 + 73.33(3.333) = 122.22 ft.
Thus the car has travelled:
d = x(3.333) - x(0) = 122.22 ft - 0 ft = 122.22 ft
-Dan
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