Position vector of a particle is given as r - 43 t32- - t22- - 2t k- m- Magnitude of acceleration at t-1 sec is-

R = [ 43 t^32î t^22ĵ + 2t k̂ ] m v =drdt = [43×32 t^12î t ĵ + 2k̂ nbsp; ]ms^ 1 =[2 t^12î t ĵ+2k̂] ms^ 1 a =dvdt = [ 2 ×12 t^ 12î ĵ] ms^ 2 a|_t=1 = î ...

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Position vector of a particle is given as $\to r = ⎛ ⎜ ⎝ \frac{4}{3} t^{\frac{3}{2}}^i - \frac{t^{2}}{2}^j + 2 t^k ⎞ ⎟ ⎠ m$ . Magnitude of acceleration at $t = 1 sec$ is:

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Solution

$\to r = ⎡ ⎢ ⎣ \frac{4}{3} t^{\frac{3}{2}}^i - \frac{t^{2}}{2}^j + 2 t^k ⎤ ⎥ ⎦ m$
$\to v = \frac{d \to r}{d t}$
$= ⎡ ⎢ ⎣ \frac{4}{3} \times \frac{3}{2} t^{\frac{1}{2}}^i - t^j + 2^k ⎤ ⎥ ⎦ {ms}^{- 1}$
$= [2 t^{\frac{1}{2}}^i - t^j + 2^k] {ms}^{- 1}$
$\to a = \frac{d \to v}{d t}$
$= ⎡ ⎢ ⎣ 2 \times \frac{1}{2} t^{- \frac{1}{2}}^i -^j ⎤ ⎥ ⎦ {ms}^{- 2}$
$\to a |_{t = 1} =^i -^j$
$| \to a | = \sqrt{1^{2} + 1^{2}}$
$= \sqrt{2} {ms}^{- 2}$

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Position vector of a particle is given as →r=⎛⎜⎝43 t32^i−t22^j+2t^k⎞⎟⎠m. Magnitude of acceleration at t=1 sec is:

→r=⎡⎢⎣43 t32^i−t22^j+2t^k⎤⎥⎦ m →v=d→rdt =⎡⎢⎣43×32 t12^i−t^j+2^k⎤⎥⎦ms−1 =[2 t12^i−t^j+2^k] ms−1 →a=d→vdt =⎡⎢⎣2×12 t−12^i−^j⎤⎥⎦ ms−2 →a|t=1=^i−^j |→a|=√12+12 =√2 ms−2

Position vector of a particle is given as $\to r = ⎛ ⎜ ⎝ \frac{4}{3} t^{\frac{3}{2}}^i - \frac{t^{2}}{2}^j + 2 t^k ⎞ ⎟ ⎠ m$ . Magnitude of acceleration at $t = 1 sec$ is: