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

In summary: Therefore$$v(t) = v(t_1) + \int_{t_1}^t a(\tau)\; d\tau $$If you insert the formulas for the aceleration ##a(\tau)## with the two different expressions for ##a(\tau)## in the two time intervals you get the ##v(t)## expressed with two pieces. In the first interval ##a=0## and you get a constant velocity (from ##t_0## up to ##t_1##), in the second interval you can integrate with the variable limit (from ##t_1## up to ##t##) and you get the second pice of the ##v(t)## function.
  • #1
fmiren
13
1
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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  • #2
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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  • #3
Yes, I see this now, what confuses me is how to choose the limits of definite integral in such kind of problems.
 
  • #4
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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Related to Finding Velocity of Car Slowing Down: v_0-c*(t-t_1)^2/2

1. What is the equation for finding the velocity of a car slowing down?

The equation for finding the velocity of a car slowing down is v_0-c*(t-t_1)^2/2, where v_0 is the initial velocity, c is the deceleration constant, t is the time, and t_1 is the time at which the car began slowing down.

2. How do you determine the initial velocity of a car?

The initial velocity of a car can be determined by measuring the car's speed at the beginning of its motion or by using a speedometer.

3. What does the deceleration constant represent in the equation?

The deceleration constant, c, represents the rate at which the car slows down. It is measured in units of distance per time squared (e.g. meters per second squared).

4. How does the time at which the car began slowing down affect the velocity?

The time at which the car began slowing down, t_1, affects the velocity because it is used to calculate the change in time, (t-t_1), which is squared and multiplied by the deceleration constant. This value is then subtracted from the initial velocity, resulting in a decrease in the overall velocity.

5. Can this equation be used to find the velocity of a car at any point during its motion?

No, this equation can only be used to find the velocity of a car at a specific point in time when it is slowing down. It cannot be used to find the velocity at any other point during the car's motion.

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