Question

At t = 0, a particle leaves the origin with a velocity of 5.0 m/s in the positive y direction. Its acceleration is given by = (3.0 - 2.0) m/s2. At the instant the particle reaches its maximum y coordinate how far is the particle from the origin?

Answer 1

Answer:

d = 11.27 m

Explanation:

initial velocity of the particles at origin is given as

$v_i = 5\hat j$

acceleration is given as

$a = 3\hat i - 2\hat j$

now when particle will reach to maximum Y coordinate then its velocity in Y direction becomes zero

$v_y = v_{iy} + a_y t$

$0 = 5 - 2t$

t = 2.5 s

now we have

$y = v_{iy} t + \frac{1}{2}a_y t^2$

$y = 5(2.5) - \frac{1}{2}(2)(2.5^2)$

$y = 6.25 m$

Similarly for x coordinate we have

$x = v_{ix} t + \frac{1}{2}a_x t^2$

$x = 0(2.5) + \frac{1}{2}(3)(2.5^2)$

$x = 9.375 m$

Now the distance from origin is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{6.25^2 + 9.375^2} = 11.27 m$

Answer 2

The distance of the particle from the origin after reaching its maximum is; 11.267m

What is the Resultant Vector Distance?

We are given;

Initial velocity; u = 5j (because it is in the positive y direction)

Acceleration; a = 3i - 2j

At maximum point, final velocity is zero. Using newton's first equation of motion in the y direction gives;

v = u + at

0 = 5j + t(-2j)

t = 5j/2j

t = 2.5 secs

Using newtons third equation of motion, we can find the distance in the y-direction as;

y = ut + ¹/₂at²

y = 5(2) + (¹/₂ * -2 * 2.5²)

y = 6.25 m

Distance in the x-direction is;

x = (0 * 2.5) + ¹/₂ * 3 * 2.5²

x = 9.375 m

Thus, distance from the origin is;

d = √(x² + y²)

d = √(6.25² + 9.375²)

d = 11.267m

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