A particle is moving with the given data. Find the position of the particle. a(t) = 2t + 5, s(0) = 7, v(0) = −7

Question

A particle is moving with the given data. Find the position of the particle.

a(t) = 2t + 5,   s(0) = 7,    v(0) = −7

Answer 1

I'm assuming that you are looking for the position of a particle on any given time t.

What you have to do is to integrate a(t) twice:

a(t) = 2t + 5

v(t) = ∫a(t) dt = ∫(2t + 5) dt = t² + 5t + C₁

Given: v(0) = 7,

v(0) = (0)² + 5(0) + C₁ = -7

C₁ = -7

The velocity at any given time is:

v(t) = t² + 5t - 7

s(t) = (1/3) t³ + (5/2) t² - 7t + C₂

Given: s(0) = 7

s(0) = (1/3) (0)³ + (5/2)(0)² - 7(0) + C₂ = 7

C₂ = 7

Therefore the position of the particle at any given time is:

s(t) = (1/3) t³ + (5/2) t² - 7t + 7

Answer 2

v(t) = ∫a(t)dt = t² + 5t + C₁ Since v(0) = -7, 0² + 5(0) + C₁ = -7. So, C₁ = -7

Therefore, v(t) = t² + 5t - 7

s(t) = ∫v(t)dt = (1/3)t³ + (5/2)t² - 7t + C₂

Since s(0) = 7, we have s(t) = (1/3)t³ + (5/2)t² - 7t + 7