JEE PYQ: Mathematics In Physics Question 21
Question 21 - 2019 (09 Apr Shift 2)
The position vector of a particle changes with time according to the relation $\vec{r}(t) = 15t^2;\hat{i} + (4 - 20t^2);\hat{j}$. What is the magnitude of the acceleration at $t = 1$?
(1) 40
(2) 25
(3) 100
(4) 50
Show Answer
Answer: (4)
Solution
$\vec{r} = 15t^2;\hat{i} + (4 - 20t^2);\hat{j}$ $\vec{v} = \frac{d\vec{r}}{dt} = 30t;\hat{i} - 40t;\hat{j}$ Acceleration, $\vec{a} = \frac{d\vec{v}}{dt} = 30;\hat{i} - 40;\hat{j}$ $\therefore a = \sqrt{30^2 + 40^2} = 50$ m/s$^2$
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