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# The velocity of a particle moving in the xy plane is given by  Assume t > 0. (a) What is the acceleration when t = 3 s? (b) When (if ever) is the acceleration zero? (c) When (if ever) is the velocity zero? (d) When (if ever) does the speed equal 10 m/s?

6 years ago
Given:
Velocity of the particle ,.
(a) The acceleration vector of the particle is given as:

Substitute the given value of in the equation above to have

Now we apply the property of differentiation given as:

Using the property of differentiation one can obtain the acceleration vector as:
…… (1)
It is important to note that the component of velocity vector along the unit vector is a constant value, therefore its differentiation is zero.
As a result, the acceleration vector contains component along unit vector  only.
At t = 3s , the acceleration vector is given as:

The negative shows that the particle is experiencing deceleration.
Therefore the magnitude of acceleration acting on the particle is 18 m/s2.
(b) The value of time t for which the acceleration is zero can be calculated by equating the acceleration vector given by equation (1) to zero that is:

The above condition occurs when the component of the acceleration vector along unit vector is zero.

Therefore at 0.75 s, the acceleration on the particle is zero.
(c) The value of time for which the velocity is zero can be calculated by equating the given velocity vector to zero, that is:

This is true when both the horizontal and vertical component of the velocity vector is zero. But as can be seen from the vector , the component along the unit vector is independent of time and will always be constant
Therefore the velocity vector can never be zero.
(d) The speed of the particle is defined as the magnitude of the velocity . If you write the velocity vector as say , then on comparing it with given vector gives:

The speed (say v) is given as the magnitude of velocity vector as:

Substitute the values of and to get speed as:

Also if the value of speed is 10/s, then one can equate v with the value in the equation above to obtain the time at which the particle has that speed as:

Squaring both sides

This is a quadratic equation in t whose solution can be founded using the solution of quadratic equation explained below:
For a quadratic equation given as:
at2 + bt + c = 0 …… (3)
The solutions of t is given as:

On comparing our quadratic equation of time given by (2) with the equation (3), we have

Therefore the solution of the quadratic equation given by (2) is:

In the solution above, the time t1 has a negative term in the square root, accounting for imaginary time.
The same would happen with the square root term of t2 because it involves same constants.
Therefore we infer that the particle will never have the speed equal to 10 m/s .