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In: Physics

A particle leaves the origin with an initial velocity of 4.10 m/s in the x direction,...

A particle leaves the origin with an initial velocity of 4.10 m/s in the x direction, and moves with constant acceleration ax = -1.40 m/s2 and ay = 3.80 m/s2. How far does the particle move in the x direction before turning around? What is the particle's velocity at this time? Enter the x component first, followed by the y component.

Expert Solution

Given the initial velocity components


And given the acceleration components


And so, from the kinematical equations, we get for the x component

and

Alternatively, we get

And for the particle to turn around in the x-direction, we require the particle velocity to become zero momentarily, i.e.,



And so, putting the given values, we get


And the time at which the particle turn around in the x direction, is given by


So, putting the given values, we get


The velocity at this time along the x direction is zero, i.e,

The motion in the y direction is given by

So, at the above time, we get the velocity in the y direction

genius_generous answered 6 months ago

A particle leaves the origin with initial velocity v⃗ =15i^+18j^m/s v→o=15i^+18j^m/s, undergoing constant acceleration a⃗ =−1.5i^+0.31j^m/s2a→=−1.5i^+0.31j^m/s2....

A particle leaves the origin with initial velocity v⃗ =15i^+18j^m/s v→o=15i^+18j^m/s, undergoing constant acceleration a⃗ =−1.5i^+0.31j^m/s2a→=−1.5i^+0.31j^m/s2. in what direction is it moving ?

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