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.

Printable View

- Apr 19th 2007, 03:41 PMUMStudentcalc 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. - Apr 19th 2007, 03:48 PMThePerfectHacker
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. - Apr 19th 2007, 03:53 PMtopsquark
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