Question

Can anyone please help me out with the question no 19?

Answer 1

$\Large\underline{\underline{\sf Given:}}$

• Position of Particle r = $\sf{6t\hat{i}+4t^2\hat{j}+10\hat{k}}$

$\Large\underline{\underline{\sf To\:Find:}}$

• Velocity and Acceleration.
• Magnitude and direction of Velocity at time 2s.

$\Large\underline{\underline{\sf Solution:}}$

(a) Velocity & Acceleration

Velocity :

$\Large\implies{\sf v=\dfrac{dr}{dt}}$

$\implies{\sf \red{v= 6\hat{i}+8t\hat{j}m/s}}$

Acceleration :

$\Large{\sf a=\dfrac{dv}{dt}}$

$\implies{\sf \red{a=8\hat{j}m/s^2 }}$

(b) Magnitude and Direction of Velocity

at time t = 2s

Velocity (v) = (6î + 16j ) m/s

Magnitude of Velocity

$\implies{\sf \sqrt{6^2+16^2} }$

$\implies{\sf \sqrt{36+256} }$

$\implies{\sf \sqrt{292} }$

$\implies{\sf \red{≈17.08\:m/s }}$

Direction ∅ with x-axis

$\implies{\sf \theta=tan^{-}\left(\dfrac{16}{6}\right)}$

$\implies{\sf \red{\theta=tan^{-1}\left(\dfrac{8}{3}\right)}}$

$\Large\underline{\underline{\sf Answer:}}$

⛬ (a) Velocity = (6î + 8t j ) m/s

Acceleration = 8j m/s²

(b) Magnitude of Velocity = 17.08m/s

Direction = ∅ = $\sf{tan^{-1}\left(\dfrac{8}{3}\right)}$

Answer 2

r =6ti + 4t^2j +10k

a. when we differntiate the position vector we will get the velocity.

Now, dr/dt=6i +8tj

velocity= 6i +8tj

when we diffrentiate the velocity with respect to time we will get accleration.

dv/dt=8j

a=8j

b. velocity at t=2

v= 6 i +8×2j

v=6i +16j.