Position vector of a particle is given as r-4-3 t3-2-t2-2-2 t k- m- Magnitude of acceleration at t -1 sec is-A- 1 m - s 2B- 2 m - s 2C- -2 m - s 2D- -5 m - s 2

The correct option is C √(2) m/s^2r = [ 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 = [ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Magnitude and Direction of Acceleration

Position vect...

Question

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:

1 {m/s}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 {m/s}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{2} {m/s}^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\sqrt{5} {m/s}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. 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:

Q. A particle is moving on a circular path of radius

1 m

and its speed is changing as

v = 2 t

. The magnitude of the acceleration of particle at time

t = 1 s

Q. A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 5 m/s and the speed is increasing at a rate of

2 m / s^{2}

. The magnitude of net acceleration at this instant is:

Q. A particle is revolving in a circular path having radius

‘ r^{'} = 1 m

at a speed of

2 m / s

. It is increasing its speed at the rate of

3 m / s^{2}

. The magnitude of total acceleration of the particle at that instant is

Q. Magnitude of acceleration of the two blocks is :
(Radius of pulley is

1 m

)

Join BYJU'S Learning Program

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:

The correct option is C √2 m/s2→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: