The velocity function is v(t)-t^2+3t-2 for a particle moving along a line. What is the displacement (net distance covered) of the particle during the time interval [-3,6]?

crubats4b5p

crubats4b5p

Answered question

2023-03-25

The velocity function is v ( t ) = t 2 + 3 t 2 for a particle moving along a line. What is the displacement (net distance covered) of the particle during the time interval [-3,6]?

Answer & Explanation

reabatebhyk

reabatebhyk

Beginner2023-03-26Added 6 answers

The area under a velocity curve equals the distance traveled.
- 3 6 v ( t ) d t
= - 3 6 - t 2 + 3 t - 2 X d t
= - 1 3 t 3 + 3 2 t 2 - 2 t ( - 3 ) 6
= ( - 1 3 ( 6 3 ) + 3 2 ( 6 2 ) - 2 ( 6 ) ) - ( - 1 3 ( - 3 ) 3 + 3 2 ( - 3 ) 2 - 2 ( - 3 ) )
= 114 - 10.5
= 103.5
Thus, - 3 6 v ( t ) d t = 103.5
inpuctists8f5

inpuctists8f5

Beginner2023-03-27Added 5 answers

The sum of the partial integrals gives the total distance (scalar quantity representing actual path length).
x = - 3 1 ( 0 - ( - t 2 + 3 t - 2 ) d t + 1 2 ( - t 2 + 3 t - 2 ) d t + 2 6 ( t 2 - 3 t + 2 ) d t
The magnitude of total displacement (vector quantity representing straight line drawn from start to end of motion) is given by the following integral.
| x | = - - 3 1 ( t 2 - 3 t + 2 ) d t + 1 2 ( - t 2 + 3 t - 2 ) d t - 2 6 ( t 2 - 3 t + 2 ) d t

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