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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) \( -1 ms ^{-2} \) |
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Answer» 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 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. |
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