The acceleration function a(t) (in m/s2) and the initial velocity v(0) are given for a particle?

Question

The acceleration function a(t) (in m/s2) and the initial velocity v(0) are given for a particle moving along a line.

a(t) = 2t + 3 0 <= t <= 3

v(0) = -4

v(t) =

total distance = Is this the absolute values of v(t)?

Please explain.

Answer 1

&lt;&lt; total distance = Is this the absolute values of v(t)? &gt;&gt;

NO ... this is not so.

GIVEN that

a(t) = 2t + 3

you need to find v(t) first before you can do any distance calculations.

First step -- integrate the above function for a(t) to determine the velocity.

a(t) = dv/dt = 2t + 3

v(t) = Integral (2t + 3)dt

v(t) = t^2 + 3t + C

where

C = constant of integration

Since at t = 0, v(0) = -4, then the equation for the velocity is modified to

v(t) = t^2 + 3t - 4

Step 2 -- integrate the above v(t) function to determine the distance travelled by the particle

v(t) = ds/dt = t^2 + 3t - 4

s = Integral (t^2 + 3t - 4)dt

s = (1/3)t^3 + (3/2)t^2 - 4t + C

where C = constant of integration

and in the absence of any other given data, it will be assumed that at

t = 0, s = 0, hence C = 0.

Thus being said,

Total distance = s = (1/3)t^3 + (3/2)t^2 - 4t

Hope this helps.

  

Answer 2

v(t) = ∫a(t)dt = ∫(2t + 3)dt = t² + 3t + C

v(0) = 0² + 3(0) + C = -4 → C = -4

v(t) = t² + 3t - 4

Believe it or not, you just solved a special kind of differential equation called an &quot;initial value problem&quot; (if you understood what I wrote).