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A Particle had a velocity of 18 m/s in the +x direction and 2.4 s later its velocity was m/s in the opposite direction. What was the average acceleration of the particle during this 2.4-s interval?

A Particle had a velocity of 18 m/s in the +x direction and 2.4 s later its velocity was m/s in the opposite direction. What was the average acceleration of the particle during this 2.4-s interval?

Grade:11

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago
We assume that the unit vector \widehat{i}is along the positive x axis whereas the unit vector\widehat{j} is along the positive y axis.
Given:
Initial velocity of the of the particle,\overrightarrow{v}_{i} = 18 m / s\ \widehat{i} .
Final velocity of the particle, \overrightarrow{v}_{f} = -30 m / s\ \widehat{i}.
Time interval over which the velocity vector of particle changes, \Delta\ t = 2.4\ s. .
It is important to note that the negative sign in the final velocity is due to the fact that the velocity vector of the particle is in a direction opposite to the unit vector \widehat{i}.
The average acceleration\overrightarrow{a}_{av} of the particle is defined as:
\overrightarrow{a}_{av} = \frac{\Delta\overrightarrow{v}}{\Delta t}…… (1)
Where\Delta \overrightarrow{v} is the change in velocity vector of the particle.
The change in the velocity vector of the particle is given as:
\Delta \overrightarrow{v} = \overrightarrow{v}_{f} - \overrightarrow{v}_{i}
Substitute the given values of the vector in the equation above to obtain as:

236-1829_33.JPG
Substitute the value of \Delta \overrightarrow{v}in equation (1) to have

236-1440_Capture.JPG
Substitute the given value of \Delta tin the equation above


236-1430_Capture2.JPG
The negative sign indicates that the particle is experiencing the acceleration in a direction opposite to the direction of unit vector \widehat{i}.
Therefore the magnitude of average acceleration of the particle is 20 m /s2.

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