Question

the position vector is given by r= 5t^2icap + 2t^3jcap + 2kcap. find the velocity and acceleration at t= 2s. pls tell me the answer and wrong answers will be reported

Answer 1

Answer:

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zero

45∘

90∘

180∘

Answer :

C

Solution :

The position ………… (V→)=dr→dt=2iˆ+10tjˆ

a→=10jˆ

V→(t=0)=2iˆ

a→(t=0)=10jˆ

⇒V→⊥a→(θ=90∘)

Answer 2

Answer:

$v=20i+24j\\a=10i+24j$

Explanation:

$r= 5t^{2}i +2t^{3}j+2k$

$v=\frac{dr}{dt}$

$v=d\frac{5t^2+2t^3+2k}{dt}$

$\frac{d}{dt} [{5t^2i}] +\frac{d}{dt} [2t^3j}]+\frac{d}{dt} [{2k}]$

$5*2t^{1}i+2*3t^{2}j+0k$

$=10ti+6t^2j$

To find velocity,v

Velocity at t=2

$(10*2)i+(6*2^{2})j\\=(20)i+(24)j$

To find acceleration,a

$a=\frac{dv}{dt}$

$\frac{d}{dt} [{(10t)i}] +\frac{d}{dt} [(6t^2)j}]$

$=(10)i+(6*2t)j\\=(10)i+(12t)j$

acceleration at t=2

$=(10)i+(12*2t)j\\=10i+24j$

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