Find the velocity and position vectors of a particle with acceleration a(t)=10k, and initial conditions v(0)= -3j-2k and r(0)=-4i-2j+5kA. v(t)= _____i+_____j+______k B. r(t)= _____i+_____j+______k

To find the velocity and position vectors of a particle, we need to integrate the acceleration vector and apply the initial conditions.

A. Velocity vector v(t)

The acceleration vector a(t) is given as 10k.

We integrate this to find the velocity vector:

∫a(t) dt = ∫10k dt = 10∫dt = 10t + C

We know that the initial velocity v(0) = -3j - 2k.

So, C = -3j - 2k.

Therefore, the velocity vector v(t) = 10t k - 3j - 2k = -3j + (10t - 2)k

B. Position vector r(t)

We integrate the velocity vector to find the position vector:

∫v(t) dt = ∫(-3j + (10t - 2)k) dt = -3∫dt j + ∫(10t - 2)k dt = -3t j + (5t^2 - 2t)k + D

We know that the initial position r(0) = -4i - 2j + 5k.

So, D = -4i - 2j + 5k.

Therefore, the position vector r(t) = -4i - 3t j + (5t^2 - 2t + 5)k.

To find the velocity and position vectors of a particle, we need to integrate the acceleration function and apply the initial conditions.

A. Velocity vector v(t)

The acceleration vector a(t) is given as 10k.

We integrate this function with respect to time t to get the velocity function v(t):

∫a(t) dt = ∫10k dt = 10∫dt k = 10t k + C

We know that the initial velocity v(0) = -3j - 2k.

This means that C = -3j - 2k.

So, the velocity function v(t) = -3j + (10t - 2)k

B. Position vector r(t)

We integrate the velocity function v(t) to get the position function r(t):

∫v(t) dt = ∫(-3j + (10t - 2)k) dt = -3∫dt j + ∫(10t - 2)k dt = -3t j + (5t^2 - 2t)k + D

We know that the initial position r(0) = -4i - 2j + 5k.

This means that D = -4i - 2j + 5k.

So, the position function r(t) = -4i - 3t j + (5t^2 - 2t + 5)k.

To find the velocity and position vectors of a particle, we need to integrate the acceleration vector and apply the initial conditions.

A. Velocity vector v(t)

The acceleration vector a(t) is given as 10k.

We integrate this with respect to time t to get the velocity vector v(t):

∫a(t) dt = ∫10k dt = 10∫dt k = 10t k + C

We know that the initial velocity v(0) = -3j - 2k.

So, C = -3j - 2k.

Therefore, the velocity vector v(t) = -3j + (10t - 2)k.

B. Position vector r(t)

We integrate the velocity vector v(t) with respect to time t to get the position vector r(t):

∫v(t) dt = ∫(-3j + (10t - 2)k) dt = -3∫dt j + ∫(10t - 2) dt k = -3t j + (5t^2 - 2t) k + D

We know that the initial position r(0) = -4i - 2j + 5k.

So, D = -4i + 5k.

Therefore, the position vector r(t) = -4i - 3t j + (5t^2 - 2t + 5)k.