Answer:
Mistake in question
The correct question
A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10t , where a and t are in m/s² and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v², where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.
Explanation:
Given the function
a = 50 —10t
The car started from rest u = 0
And it accelerates to a speed of 125m/s
Then, let find the time in this stage
Acceleration can be modeled by
a = dv/dt
Then, dv/dt = 50—10t
Using variable separation to solve the differentiation equation
dv = (50—10t)dt
Integrating both sides
∫ dv = ∫ (50—10t)dt
Note, v ranges from 0 to 125seconds, so we want to know the time when it accelerate to 125m/s. So t ranges from 0 to t'
∫ dv = ∫ (50—10t)dt
v = 50t —10t²/2. Equation 1
[v] 0<v<125 = 50t —10t²/2 0<t<t'
125—0 = 50t — 5t² 0<t<t'
125 = 50t' — 5t'²
Divide through by 5
25 = 10t' — t'²
t'² —10t' + 25 = 0
Solving the quadratic equation
t'² —5t' —5t' + 25 = 0
t'(t' —5) —5(t' + 5) = 0
(t' —5)(t' —5) = 0
Then, (t' —5) = 0 twice
Then, t' = 5 seconds twice
So, the car spent 5 seconds to get to 125m/s.
The second stage when the parachute was deployed
We want to the time parachute reduce the speed from 125m/s to 10m/s,
So the range of the velocity is 125m/s to 10m/s. And time ranges from 0 to t''
The function of deceleration is give as
a = - 0.02v²
We know that, a = dv/dt
Then, dv/dt = - 0.02v²
Using variable separation
(1/0.02v²) dv = - dt
(50/v²) dv = - dt
50v^-2 dv = - dt
Integrate Both sides
∫ 50v^-2 dv = -∫dt
(50v^-2+1) / (-2+1)= -t
50v^-1 / -1 = -t
- 50v^-1 = -t
- 50/v = - t
Divide both sides by -1
50/v = t. Equation 2
Then, v ranges from 125 to 10 and t ranges from 0 to t''
[ 50/10 - 50/125 ] = t''
5 - 0.4 = t''
t'' = 4.6 seconds
Then, the time taken to decelerate from 125s to 10s is 4.6 seconds.
So the total time is
t = t' + t''
t = 5 + 4.6
t = 9.6 seconds
b. Total distanctraveleded.
First case again,
We want to find the distance travelled from t=0 to t = 5seconds
a = 50—10t
We already got v, check equation 1
v = 50t —10t²/2 + C
v = 50t — 5t² + C
We add a constant because it is not a definite integral
Now, at t= 0 v=0
So, 0 = 0 - 0 + C
Then, C=0
So, v = 50t — 5t²
Also, we know that v=dx/dt
Therefore, dx/dt = 50t — 5t²
Using variable separation
dx = (50t —5t²)dt
Integrate both sides.
∫dx = ∫(50t —5t²)dt
x = 50t²/2 — 5 t³/3 from t=0 to t=5
x' = [25t² — 5t³/3 ]. 0<t<5
x' = 25×5² — 5×5³/3 —0
x' = 625 — 208.333
x' = 416.667m
Stage 2
The distance moved from
t=0 to t =4.6seconds
a = -0.002v²
We already derived v(t) from the function above, check equation 2
50/v = t + C.
When, t = 0 v = 125
50/125 = 0 + C
0.4 = C
Then, the function becomes
50/v = t + 0.4
50v^-1 = t + 0.4
Now, v= dx/dt
50(dx/dt)^-1 = t +0.4
50dt/dx = t + 0.4
Using variable separation
50/(t+0.4) dt = dx
Integrate both sides
∫50/(t+0.4) dt = ∫ dx
50 In(t+0.4) = x
t ranges from 0 to 4.6seconds
50In(4.6+0.4)—50In(4.6-0.4) = x''
x'' = 50In(5) —50In(4.2)
x'' = 8.72m
Then, total distance is
x = x' + x''
x = 416.67+8.72
x = 425.39m
The total distance travelled in both cases is 425.39m
