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Standard XII

Physics

Integration

At the moment...

Question

At the moment $t = 0$ particle leaves origin and moves in the positive direction of $x$ -axis. Its velocity as a function of time is given as $v = (5 - t) m/s$ . $x$ -coordinate of particle at $t = 2 s$ is

8 m

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6 m

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10 m

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12 m

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Solution

The correct option is A 8 m
$v = \frac{d x}{d t} = (5 - t) \Rightarrow \int_{0}^{x} d x = \int_{0}^{2} (5 - t) d t$
$x = [5 t - \frac{t^{2}}{2}]_{0}^{2} \Rightarrow x = 5 \times 2 - \frac{2^{2}}{2} = 8 m$

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At the moment t=0 particle leaves origin and moves in the positive direction of x-axis. Its velocity as a function of time is given as v=(5−t) m/s. x-coordinate of particle at t=2 s is

The correct option is A 8 mv=dxdt=(5−t)⇒∫x0dx=∫20(5−t)dt x=[5t−t22]20⇒x=5×2−222=8 m

At the moment $t = 0$ particle leaves origin and moves in the positive direction of $x$ -axis. Its velocity as a function of time is given as $v = (5 - t) m/s$ . $x$ -coordinate of particle at $t = 2 s$ is

The correct option is A 8 m
$v = \frac{d x}{d t} = (5 - t) \Rightarrow \int_{0}^{x} d x = \int_{0}^{2} (5 - t) d t$
$x = [5 t - \frac{t^{2}}{2}]_{0}^{2} \Rightarrow x = 5 \times 2 - \frac{2^{2}}{2} = 8 m$