Problem 9: A detailed graph of position versus time is...

Problem 9: A detailed graph of position versus time is shown. x(km) 30.0 20.0 10.0 1.0 2.0 3.0 4.0 5.0 t(h) 20% Part (a) What is the maximum displacement represented on the graph in km? $x_{max} = 30.00$ km ? Correct! 20% Part (b) What is the acceleration at $t = 1.1$ h in km/h²? $a = 0.000$ km/h² ? Correct! 20% Part (c) What is the instantaneous velocity at $t = 2.6$ h in km/h? $v = -15$ $v = -15.00$ km/h ? Correct! $v = $ 20% Part (d) What is the average velocity during the time interval from 1.65 h to 4.3 in km/h? ? 20% Part (e) What is the average position during the interval from 0.15 h to 4.3 h in km? $\bar{x} = $ km

Transcript

00:02 Hello, in the question we have given this distance -time graph.

00:07 So, this distance is in kilometers and this time is in hours.

00:12 So, what we have to do is we have to solve the following parts.

00:17 So, the part a is what is the maximum displacement represented on the graph in kilometers.

00:26 So, the maximum displacement is so x maximum as mentioned it is 30 kilometers.

00:38 Then part b is what is the acceleration, what is the acceleration at t is equal to 1 .1 hour.

00:51 So, time is given that is 1 .1 hour.

00:56 So, now if we see 1 .1 will be somewhere over here and this is a constant velocity.

01:05 So, constant velocity means acceleration will be equal to zero.

01:09 Now coming to part c.

01:11 So, part c is what is the instantaneous velocity at t is equal to 2 .6 kilometer per hour.

01:17 So, t is equal to 2 .6 kilometer per hour, sorry 2 .6 hour, extremely sorry in kilometer per hour.

01:31 So, we have to find out the instantaneous velocity.

01:34 So, now this will be 30 minus 0.

01:39 So, this v instantaneous will be 30 minus 0 sorry final is 0 minus initial is 30 divided by 2 by because this is 4 minus 2 4 minus 2.

01:56 So, this gives us 30 by 2 that is equals to minus 15 kilometer per hour.

02:08 So, now this will be the instantaneous speed because over here the speed is constant instantaneous velocity.

02:13 So, the instantaneous velocity will be minus 15 kilometer per hour.

02:18 Now the part d.

02:22 So, what is the average velocity during the time interval from 1 .75 to 4 .9 in kilometer per hour.

02:37 So, part d is what is average velocity during the time interval.

02:44 So, time interval is given that is t is equal to 1 .75 hours to t is equal to 4 .9 hours in kilometer per hour.

02:56 Now if we notice 1 .75.

02:59 So, it turns out to be somewhere over here.

03:01 So, this is 1 .4 1 .6.

03:04 So, this is somewhere over here.

03:06 So, it is 30.

03:07 So, over here this velocity is 30 and 4 .9 4 .9 means somewhere over here.

03:14 So, over here this is 0.

03:17 So, 30 0 minus 30.

03:20 So, v average will be equal to 0 minus 30 divided by this 4 .9 minus 1 .75...

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