# Math Help - calc problem rectilineal motion

1. ## calc problem rectilineal motion

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.

2. Originally Posted by UMStudent
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:

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.

3. Originally Posted by UMStudent
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.
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