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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?

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?

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
9 years ago
Given:
Velocity of the particle ,236-346_1.JPG.
(a) The acceleration vector of the particle is given as:
\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt}
Substitute the given value of \overrightarrow{v}in the equation above to have
236-1613_2.JPG
Now we apply the property of differentiation given as:

236-357_3.JPG
Using the property of differentiation one can obtain the acceleration vector as:
236-2445_4.JPG …… (1)
It is important to note that the component of velocity vector along the unit vector\widehat{j} is a constant value, therefore its differentiation is zero.
As a result, the acceleration vector contains component along unit vector \widehat{i} only.
At t = 3s , the acceleration vector is given as:
236-117_5.JPG
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:
236-1563_6.JPG
The above condition occurs when the component of the acceleration vector along unit vector\widehat{i} is zero.
236-1419_7.JPG
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 \overrightarrow{v}to zero, that is:
236-1210_8.JPG
This is true when both the horizontal and vertical component of the velocity vector is zero. But as can be seen from the vector\overrightarrow{v} , the component along the unit vector \widehat{j}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\overrightarrow{v} = v_{x}\widehat{i} + v_{y}\widehat{j} , then on comparing it with given vector gives:
236-157_9.JPG
The speed (say v) is given as the magnitude of velocity vector\overrightarrow{v} as:
236-343_10.JPG
Substitute the values ofv_{x} andv_{y} to get speed as:
236-2184_11.JPG
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:
236-908_11.JPG
Squaring both sides
236-862_13.JPG
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:
236-1170_14.JPG
On comparing our quadratic equation of time given by (2) with the equation (3), we have
236-482_15.JPG
Therefore the solution of the quadratic equation given by (2) is:
236-2137_16.JPG
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 .

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