Kinematic based question that may include finite integration and constant acceleration

[dsc2010]

Hello all, i am having some trouble finding the right way to solve the following question 25(ii). I assumed that i had to use finite integration as acceleration is not constant for the first part of the question but am not getting the right answer for distance AB.
i tried integrating the equation of Change of velocity but am getting a different answer compared to the book :32
Can someone please help me out?

Question:



Expected answer:



Workings:

 

[HallsofIvy]

Initially the car has a velocity of 20 m/s and then has acceleration [math]\frac{dv}{dt}= \frac{3t}{2}- 6[/math]. [math]dv= \left(\frac{3t}{2}- 6\right)dt[/math] and, integrating, the car's velocity at T additional seconds is [math]v(T)= \frac{3T^2}{4}- 6T+ 20[/math]. When T= 4, [math]v(4)= \frac{3(16)}{4]- 6(4)+ 20= 12- 24+ 20= 8[/math] m/s. Yes, that is what you have.

I don't see what work you did for the second part. To find the distance from A to B, integrate again. [math]D(T)= \frac{T^3}{4}- 3T^2+ 20T[/math] and [math]D(4)= \frac{64}{4}- 3(16)+ 20(4)= 16- 48+ 80= 48[/math] m.

Finally, the car accelerates at 2 m/s^2: [math]\frac{dv}{dt}= 2[/math] so [math]v(t)= 2t+ 8= 20[/math]. [math]2t= 12[/math], [math]t= 6[/math] seconds. Accelerating at 2 m/s^2 it will take 6 seconds to regain a speed of 20 m/s.

 

[dsc2010]

Ops, i missed the last screenshot.

Many thanks for the detailed explanation

My mistake was that i didnt include the initial velocity in the equation before integrating to get the distance covered. ( + 20 )