Solving the Particle Velocity at Time t=5

[thereddevils]

Homework Statement

At time , t s , a particle with a mass 1 kg moves under the action of force F=6(i+t j) N. If the particles starts to move at time t=0 ,find the velocity of the particle at t=5 .

Homework Equations

The Attempt at a Solution

By F=ma

the acceleration vector , a = 6i+6t j

integrate this gives , v= (6t+c)i+(3t^2+c)j

and when t=0 , v=0

therefore , v=(6t)i+(3t^2)j

and at t=5 s, v=(30)i+(75)j and my next job is to find its magnitude

Is everything all right ?

 

[rock.freak667]

Yes that looks to be correct, although you can leave your answer in the vectro form v=30i+75j. If you just put the magnitude, you'd need to put the angle as well since velocity is a vector quantity in that it has both magnitude and direction.

 

[CompuChip]

Just a tiny comment... you write F = ma and then write down a at once. You have used that the mass is equal to 1 here, you might have mentioned that more explicitly, perhaps.

Also the integrations are independent, so the "+ c" that goes with the i-component is not necessarily the same as the "+ c" in the j-component. You would better use different letters for those, for example

v= (6t+c₁)i+(3t^2+c₂)j
Then from v(t = 0) = 0 (= 0 i + 0 j) it indeed follows that c₁ = c₂ = 0 (but in general, you can get different constants).

Just some nitpicking from my side though, because you did correctly solve the problem.

 

[thereddevils]

thank you !

 

[rdl]

Your solution looks correct so far. To find the magnitude of the velocity, you can use the formula v = √(v_x^2 + v_y^2), where v_x and v_y are the x and y components of the velocity. Substituting in the values at t=5, we get v = √(30^2 + 75^2) = √(900 + 5625) = √6525 ≈ 80.74 m/s. This is the magnitude of the velocity at t=5.