A car is traveling on a highway at 67 mph (or 30 m/s). How far does the car travel from the time the brakes are applied until the car stops, if the car brakes with a deceleration of the following;

Question

Answer 1

second part is easier so try that first

distance = d = -15/2t^2 + 30t

d'= v = -15t +30 = 0

t=2 seconds when velocity is zero

plug into original equation to get d

d=-30+30 = 60 meters

first part is more tedious

d=-10/2t^1/2(t)^2 + 30t = -5t^5/2 + 30t

d' = v = 0 = -25/2t^3/2 + 30

solve for t=(12/5)^2/3 = about 1.8

plug back into the original equation to solve for

distance = -5(1.8)^2+30(1.8) = 37.8 meters

Answer 2

Molly, this is a simple application of the (Newtonian) equations of motion.

Since acceleration is a function of t, a is not constant. Therefore =, we must integrate to find the correct equations of motion.

by definition: dv/dt = a = -10√t

=> ∫ [v₀ to v] dv = ∫ [0 to t] -10t^1/2 dt replacing √t by t^1/2 for ease of integration

v - v₀ = -10 x 2/3 x t^3/2

v = v₀ - 20/3 t^3/2 . . . . . . . . . (1)

When it stops, when v =0: 0 = 30 - 20/3 t^3/2

t^3/2 = 9/2

t = (9/2)^2/3

From eqn (1), v = ds/dt = v₀ - 20/3 t^3/2 where s = distance traveled

∫ [0 to s] ds = ∫ [0 to t] (v₀ - 20/3 t^3/2) dt

s = v₀t - 20/3 x 2/5 x t^5/2

= v₀t - 8/3 t^5/2 . . . . . . . . (2)

= 30 (9/2)^2/3 - 8/3 ((9/2)^2/3)^5/2

= 30 (81/4)^1/3 - 8/3 (9/2)^5/3

= 49.1 meters

In part 2, deceleration is a constant, so we can use the standard equations of motion:

v² = (v₀)² + 2as

0 = 30² + 2(-15)s

s = -900/-30 = 30 meters

Answer 3

PART 1: ACCELERATION is a function of time

Velocity = Integral [from 0 to t] ( De-Acceleration ) dt

V = Integral [from 0 to t] ( 10 t^(1/2) ) dt

V = 10 [ (2/3) * t^(3/2) ] from 0 to t

V = (20/3) * t^(3/2)

In order to stop ...

30 = (20/3) * t^(3/2)

9/2 = t^(3/2)

(9/2)^(2/3) = t. ==> Time to Stop = (9/2)^(3/2)

D = Distance

D = Integral [from 0 to t ] ( V0 - V ) dt

D = integral [from 0 to (9/2)^(3/2) ] ( 30 - (20/3) * t^(3/2) ) dt

D = [ 30t - (20/3)*(3/5) *t^(5/3) ] from 0 to (9/2)^(3/2)

D = 30*(9/2)^(3/2) - (4)*[ (9/2)^(2/3) ]^(5/3)

D = 30*(9/2)^(3/2) - (4)*(9/2)^(10/9)

D = 81.77 - 21.274

D = 60.495 m

PART 2: CONSTANT ACCELERATION of -15 m/s^2

D = Vt + (1/2)*A*t^2

We just need the time.

Initial Velocity = De-Acceleration * time

30 = 15 * t

2 = t

D = Vt + (1/2)*A*t^2

D = 30(2) - (1/2)*(15)*(2)^2

D = 60 - (1/2)*(15)*(4)

D = 60 - 30

D = 30 m