Question

suppose the velocity of a particle traveling along the x-axis is given by v(t)= 3t^2+8t+15 m/s at time t seconds. The particle is initially located 5 meters left of the origin. How far does the particle travel from t=2 seconds to t=3 seconds? After 3 seconds, where is the particle with respect to the origin?

Answer 1

Answer:

Step-by-step explanation:

1) How far does the particle travel from t=2 seconds to t=3 seconds?

$\int\limits^3_2 {(3t^2+8t+15)} \, dt =\Bigg [ t^3+4t^2+15t\Bigg]^3_2\\\\=27+36+45-(8+16+30)\\\\=108-54\\\\=54 (m)$

2)

$\int\limits^3_0 {(3t^2+8t+15)} \, dt =\Bigg [ t^3+4t^2+15t\Bigg]^3_2=108$

After 3 seconds, where is the particle with respect to the origin?

108-5=103 (m)