Finding Velocity of Car Slowing Down: v_0-c*(t-t_1)^2/2

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A car initially moves with a constant velocity v0 until time t1, after which it begins to decelerate according to the acceleration function a(t). The correct expression for the car's velocity during the deceleration phase, from t1 to t2, is v(t) = v0 - c*(t - t1)^2/2. The confusion arises from the limits of integration when calculating the velocity from the acceleration function; the upper limit should be the current time t, while the lower limit is t1. Understanding the proper setup for the definite integral is crucial for accurately determining the car's velocity during the slowing down interval.
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Homework Statement
I don't understand the limits of the definite integral in the acceleration problem below.
Relevant Equations
Acceleration and velocity equations
At time t=0 , a car moving along the + x -axis passes through x=0 with a constant velocity of magnitude v0 . At some time later, t1 , it starts to slow down. The acceleration of the car as a function of time is given by:

a(t)= 0 0≤t≤t1
-c(t−t1) t1<t2

where c is a positive constants in SI units, and t1<t≤t2 is the given time interval for which the car is slowing down. Express your answer in terms of v_0 for v0 , t_1 for t1 , t_2 for t2 , and c as needed. What is v(t) , the velocity of the car as a function of time during the time interval t1<t≤t2?

To get the velocity I integrate the accelaeration function and get v_0-c*(t_2-t_1)^2/2 since I think these should be the boundaries of the definite integral. Bu the correct answer is v_0-c*(t-t_1)^2/2 and they integrate from t (upper limit) to t1 (lower limit).
Could you please help me to understand it?
 
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Your answer would mean that the velocity is a constant. Your expression does not satisfy
$$a = {dv\over dt} = c(t-t_1) $$since it is independent of ##t##.
I.e. your ##a =0## !
Do you agree ?
 
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Yes, I see this now, what confuses me is how to choose the limits of definite integral in such kind of problems.
 
The velocity at some moment ##t## is the integral fom some start up to that same ##t## -- so ##t## is the upper limit of the integration$$v(t)- v(t_1) = \int_{t_1}^t a(\tau)\; d\tau $$
(the integrand is just a 'dummy variable' ; I used ##\tau## to designate it).
 
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