Question

A particle moves from the point (2i^+4j^)m at t=0, with an initial velocity (5i^+4j^)m/s. It is acted upon by a constant acceleration (4i^+4j^)m/s2. What is the distance of the particle from the origin at time t=2s in meters

Answer 1

Answer:

the distance of the particle from the origin at time t=2s is 28 meters

Explanation:

As we know that particle is moving with uniform acceleration

So here we can use equation of kinematics in order to find the final position

here we will have

$r_f = r_i + v_i t + \frac{1}{2}at^2$

here we know that

$\vec r_i = 2\hat i + 4 \hat j$

$\vec v_i = 5\hat i + 4\hat j$

$\vec a = 4\hat i + 4\hat j$

so we will have

$r_f = (2\hat i + 4 \hat j) + (5\hat i + 4\hat j)(2) + \frac{1}{2}(4\hat i + 4\hat j)2^2$

$r_f = (2\hat i + 4 \hat j) + (10 \hat i + 8 \hat j) + (8\hat i + 8\hat j)$

$r_f = 20\hat i + 20 \hat j$

now distance from origin is given as

$d = \sqrt{20^2 + 20^2}$

$d = 20\sqrt2$

$d = 28 m$

#Learn

Topic : Kinematics

https://brainly.in/question/1168545

Answer 2

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