Question

Consider a car initiating its motion from a position 20 meters along a straight path. The car's initial velocity is 10 m/s, and its acceleration is given by $a(t) = (6t - 12)$ m/s$^2$. Determine the particle position, total distance traveled, average velocity, the average speed, and the acceleration of the particle when $t = 6$ s and between the period of $t = 3$ s and $t = 6$ s. 

          Consider a car initiating its motion from a position 20 meters along a straight path. The car's initial velocity is 10 m/s, and its acceleration is given by $a(t) = (6t - 12)$ m/s$^2$. Determine the particle position, total distance traveled, average velocity, the average speed, and the acceleration of the particle when $t = 6$ s and between the period of $t = 3$ s and $t = 6$ s.
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Consider a car initiating its motion from a position 20 meters along a straight path. The car's initial velocity is 10 m/s, and its acceleration is given by a(t) = (6t - 12) m/s^2. Determine the particle position, total distance traveled, average velocity, the average speed, and the acceleration of the particle when t = 6 s and between the period of t = 3 s and t = 6 s.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Consider a car initiating its motion from a position 20 meters along a straight path. The car's initial velocity is 10(m)/(s), and its acceleration is given by a(t)=(6t-12)(m)/(s^(2)). Determine the particle position, total distance traveled, average velocity, the average speed, and the acceleration of the particle when t=6s and between the period of t=3s and t=6s. Consider a car initiating its motion from a position 20 meters along a straight path. The car's initial velocity is 10 m/s, and its acceleration is given by a(t) = (6t - 12) m/s2. Determine the particle position, total distance traveled, average velocity, the average speed, and the acceleration of the particle when t=6 s and between the period of t=3 s and t= 6 s.
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Transcript

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00:01 Hello, the question is taken from physics and the question is given that the position of particle is defined in terms of time by the equation below where x is in meter t is in second determined given that x is equal to 4 t.
00:14 I'm just dividing it by t plus 1 by t plus 3 by t to the power 6 plus 4.
00:22 Okay and we know that what is the average velocity from t is equal to 1 to t is equal to 3.
00:31 Okay, so we know that a v average is equal to x2 minus x1 total distance traveled by total time taken.
00:40 Okay, total distances basically 4 into 3 to the power 5.
00:44 So that will be 4 into 3 to the power 5.
00:49 So that is 9 .7 2 plus 1 by 3 plus 1 by 3 to the power 5 plus 4 minus 4 minus 1 minus 1.
01:01 3 minus 4 divided by 3 minus 2 that is 3 minus 1 that is 2 so 4 will be cancelled out and this value will this will give you the value we average is equal to 9 7 2 plus 1 by 3 plus 1 by 3 to 2 to the power 5 and minus 8 divided by 2 so that is 4 8 8 2 .2 meter per second which is the required value of the average velocity okay i think something wrong with the statement that will be 40 square okay if we take it 40 square then it will be 9 into 4 36 so 36 6 minus 8 divided by 2 so that is close to 15 okay so i think it will be t square, no t.
02:11 So, so it t cube, no t to the power six.
02:14 You can check the exact value, but i'm solving it by taking it as t to the power six.
02:20 Instantaneous velocity is just dx over dt.
02:25 Okay.
02:26 At a particular time instant, that is 2 .5...
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