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
, where 
is the coefficient of friction and 
is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have: