Question

A particle leaves the origin with an initial velocity v=3i^+4j^ and constant acceleration a =-i^-0.5j^ when the particle reaches it's maximum x coordinate what is the y coordinate?​

a) 27/4 m
b) 37/4 m
c) 29/4 m
d) 39/4 m

Answer 1

`(39/4)`.

Step-by-step explanation:

In the vector notation in this question, the coefficient of
`\vec{i}` is the
`x`-component of the vector while the coefficient of
`\vec{j}` is the
`y`-component.

For example, in this question, the
`x`-component of acceleration would be
`(-1)`, while the
`y`-component of this acceleration would be
`(-0.5)`.

Because the the
`x`-component of acceleration is negative, the velocity of this particle would keep decreasing.

While the
`\! x`-component of the velocity of the particle was initially
`3` (which is positive,) this value would become zero and then negative after some amount of time. When that happens, the direction of the motion of this particle in the
`x\!`-component would be reversed, and the particle would not go any further in its original direction of motion.

Divide the change in velocity
`\Delta v_(x)` in the
`x\!`-component by the
`x`-component of acceleration
`a_(x)` to find the time required for the particle to reach the maximum
`x`-coordinate:

`\displaystyle (\Delta v_(x))/(a_(x)) = (0 - 3)/(1) = 3`.

In other words, the
`x`-coordinate of this particle would be maximized exactly at
`3\!` units of time.

Given that the acceleration of this particle is constant, apply the following SUVAT equation to find the position
`x` of this particle at time
`t`:

`\displaystyle y = (1)/(2)\, a_(y)\, t^(2) + u_(y)\, t + y_(0)`,

Where:

• 
  `a_(y) = -0.5` is the acceleration of this particle in the
  `y`-component,
• 
  `u_(y) = 4` is the initial velocity of this particle in the
  `y`-component, and
• 
  `y_(0) = 0` is the
  `y`-coordinate of the initial position (the origin) of this particle.

Evaluate this expression at
`t = 3` when the
`x`-coordinate of this particle is maximized:

`\begin{aligned} y &= (1)/(2)\, a_(y)\, t^(2) + u_(y)\, t + y_(0) \\ &= (1)/(2)\, (-0.5)\, (3)^(2) + 4\, (3) + 0 \\ &= (39)/(4)\end{aligned}`.

In other words, when the
`x`-coordinate of this particle is maximized, the
`y`-coordinate of this particle would be
`(39/4)`.