Question

The position of a particle is given by $ \overrightarrow{r}=2{{t}^{2}}\hat{i}+3t\hat{j}+4\hat{k}, $ where $t$ is in second and the coefficients have proper units for $ \overrightarrow{r} $ to be in metre. The $ \overrightarrow{a}(t) $ of the particle at $ t=1\text{ }s $ is

• A. $ 4\text{ }m{{s}^{-2}} $ along $ \text{y-} $ direction
• B. $ \text{3 }m{{s}^{-2}} $ along $ x\text{-} $ direction
• C. $ 4\text{ }m{{s}^{-2}} $ along $ x\text{-} $ direction
• D. $ \text{2 }m{{s}^{-2}} $ along $ \text{z-} $ direction
• E. $ \text{3 }m{{s}^{-2}} $ along $ \text{z-} $ direction

Answer 1

Answer: (C)

Distance, $ \overrightarrow{\mathbf{r}}=2{{t}^{2}}\mathbf{\hat{i}}+3t\mathbf{\hat{j}}+4\mathbf{\hat{k}} $
Velocity, $ \overrightarrow{\mathbf{v}}=\frac{\overrightarrow{\mathbf{dr}}}{dt}=4t\mathbf{\hat{i}}+3\mathbf{\hat{j}} $
Acceleration, $ a=\frac{dv}{dt}=4\mathbf{\hat{i}} $
$ =4\text{ }m/{{s}^{2}} $ in the $ x- $ direction.