A particle starts from the origin at t = 0 with an initial velocity of 4.8m/s along the positive x axis.If the acceleration is (-2.5 i^ + 4.3 j^)m/s2, determine (a)the velocity and (b)position of the particle at the moment it reaches its maximum x coordinate

Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Why did the particle become an aspiring gymnast? Because it wanted to reach its maximum x-coordinate! Let's start solving this physics problem and see where it takes us.

First, let's find the velocity of the particle. We know that velocity is the derivative of position with respect to time. So, to find the velocity, we'll integrate the acceleration with respect to time.

(a) Integrating the x-component of the acceleration:
∫(-2.5) dt = -2.5t

Integrating the y-component of the acceleration:
∫4.3 dt = 4.3t

So, the velocity will be:
v = (4.8 - 2.5t) i^ + (4.3t) j^

Now, let's find the time it takes for the particle to reach its maximum x-coordinate. The maximum x-coordinate occurs when the velocity in the x-direction is zero. So, we'll set the x-component of the velocity to zero and solve for t:

4.8 - 2.5t = 0
2.5t = 4.8
t = 1.92 seconds

(b) Now that we have the time, we can substitute it back into the equation for the velocity to find the velocity at that moment:

v = (4.8 - 2.5(1.92)) i^ + (4.3(1.92)) j^
v = -0.36 i^ + 8.256 j^

Finally, let's find the position of the particle when it reaches its maximum x-coordinate. To do this, we'll integrate the velocity with respect to time:

∫(-0.36) dt = -0.36t

∫8.256 dt = 8.256t

So, the position will be:
r = (-0.36t) i^ + (8.256t) j^

Substituting t = 1.92, we get:
r = (-0.36(1.92)) i^ + (8.256(1.92)) j^
r = (-0.6912) i^ + (15.8592) j^

Therefore, the position of the particle at the moment it reaches its maximum x-coordinate is approximately -0.69 i^ + 15.86 j^.

Remember, physics can be serious, but it's always good to have a little humor to keep things bouncing!

To find the velocity and position of the particle when it reaches its maximum x coordinate, we need to integrate the acceleration function.

(a) To find the velocity, we integrate the acceleration function with respect to time:

v(t) = ∫((-2.5 i^) + (4.3 j^)) dt

Since the initial velocity is along the positive x-axis, the y-component of the acceleration does not affect the x-component of the velocity.

∫((-2.5 i^) + (4.3 j^)) dt = -2.5t i^ + C

The constant C represents the initial velocity, which is 4.8 m/s.

Therefore, the x-component of the velocity is given by:

v(x) = -2.5t + 4.8

(b) To find the position, we integrate the x-component of the velocity function with respect to time:

∫(-2.5t + 4.8) dt = -1.25t^2 + 4.8t + D

The constant D represents the initial position, which is the origin (0,0).

The maximum x-coordinate is reached when the y-component of the velocity becomes 0. Since the y-component of the velocity is not affected by the x-component of the acceleration, it remains constant:

v(y) = 4.3 j^

To find the time when the y-component of the velocity becomes 0, we set v(y) equal to 0:

4.3 = 0

The y-component of the velocity is always non-zero, so the time at which the maximum x-coordinate is reached is when t = 0.

Plugging this into the position function, we get:

x = -1.25(0)^2 + 4.8(0) + 0 = 0

Therefore, the position of the particle at the moment it reaches its maximum x-coordinate is (0, 0).

In summary, the answers are:
(a) The velocity is -2.5t + 4.8 m/s.
(b) The position is (0, 0).

To determine the velocity and position of the particle at the moment it reaches its maximum x-coordinate, we need to integrate the acceleration to find the velocity and then integrate the velocity to find the position.

Given:
Initial velocity u = 4.8 m/s
Acceleration a = -2.5 i^ + 4.3 j^ m/s²

(a) Velocity (v):
To find the velocity, we'll integrate the acceleration with respect to time.
v = ∫a dt

Since the acceleration only has an x-component, we can ignore the y-component for now.
∫a dt = ∫(-2.5 i^) dt = -2.5 ∫dt = -2.5t + C₁

C₁ is the integration constant, which represents the initial velocity at t = 0. Since the initial velocity is given as 4.8 m/s along the positive x-axis, C₁ = 4.8 m/s.

v = -2.5t + 4.8

(b) Position (r):
To find the position, we'll integrate the velocity with respect to time.
r = ∫v dt

Since the initial position is the origin (0,0), the position vector r = xi^ + yj^.
∫v dt = ∫(-2.5t + 4.8) dt = (-2.5/2)t² + 4.8t + C₂

C₂ is the integration constant, which represents the initial position at t = 0. Since the initial position is at the origin, C₂ = 0.

r = (-2.5/2)t² + 4.8t

To find the maximum x-coordinate, we need to find the time when the particle reaches maximum displacement along the x-axis, i.e., when the velocity (v) becomes zero.

-2.5t + 4.8 = 0
Solving for t:
t = 4.8 / 2.5 = 1.92 seconds

Now we can substitute this time (t = 1.92) into the position equation to find the maximum x-coordinate.

r = (-2.5/2)(1.92)² + 4.8(1.92)
r = -2.352 + 9.216
r = 6.86 meters

Therefore, the velocity at the maximum x-coordinate is 0 m/s, and the position at the maximum x-coordinate is 6.86 meters.

lets just look at x direction..

d(t)=vix*t+1/2 a t^2

dx(t)=4.8t-1/2 *2.5*t^2
now to find the max
d'/dt=0=4.8-2.5 t or max occurs at
time=4.8/2.5
now solve for position at this time:
d(t)= (4.8*t -1/2 *2.5*t^2) i^ + 4.3t^2 j^

and finally, velocity (derivitative of d(t)

v(t)=(4.8 -2.5t)i^ +8.3t j^