Question

If the velocity is \( \vec{v}=2 \hat{i}+t^{2} \hat{j}-9 \hat{k} \), then the magnitude of acceleration at \( t=0.51 \) is- 

(a) \( 1 ms ^{-2} \) 

(b) \( 2 ms ^{-2} \) 

(c) zero 

(d) \( -1 ms ^{-2} \)

Answer 1

Correct option is (a) 1.02m/s²

Given \(\vec V=2\hat i+t^2\hat j-9\hat k\) 

Acceleration a = \(\frac{d\vec v}{dt}\) 

a = \(\frac d{dt}\)(2 + t² - 9)

a = 2t

\(\because\) t = 0.51 sec

a = 2 x 0.51

a = 1.02 m/s²

Answer 2

It is given that \(\vec V = 2\hat i + t^2\hat j-9\hat k\)

Velocity in the x direction = 2 m/s ⇒ not changing with respect to time

Therefore, acceleration in x direction is zero. 

Similarly, the acceleration in z direction is also zero. 

In y direction, the velocity is t² ⇒ acceleration,

 \(a = \dfrac{dv}{dt} = \dfrac{d}{dt}{(t^2)} = 2t = 2\times 0.51 = 1.02~\text{m/s}^2\)

The resultant acceleration will be approximately 1 m/s² since the acceleration in other directions is zero. 

Therefore, option A is correct.