0:00
This is the given question.
00:01
In 1967, new zealander buttmoor was set the world record for indian motorcycle with a maximum speed of 182 .59 miles per hour.
00:12
So he reached this as the maximum speed.
00:16
And the one -way course was 5 .7 miles long.
00:21
The acceleration rates are more often described by the time it takes to reach 60 miles per hour from rest.
00:28
And the time it took was 4 .2 seconds and butt accelerated at this rate until he reached his maximum speed after that he traveled with the same speed for the remaining journey that mean the statement means that and we asked how long did it take but to complete the course so we are asked good to find the time here i represent the question in a diagrammatical form so you can understand it very easily suppose consider this is the track this is the one -way course which is 5 .7 miles long suppose this is the starting point yes and this is the sorry suppose the starting point is a and the end point is b so the whole course is this is 5 .7 miles long so here he starts traveling here he starts traveling with initial velocity u equal to zero okay and the accelerate at a rate given, see, we are given the acceleration rate that it reads 60, so we have to calculate the acceleration rate and it was conistent, see, but accelerated at this rate, so it is constant until he reaches maximum speed.
01:58
Suppose consider at some point, some intermediate point c.
02:03
Suppose at intermediate point c here he reaches to a maximum speed v which is equal to 182 .59 miles per hour so this is maximum speed we he reached and after that he traveled with the same speed for the remaining journey so for the remaining journey sorry so for the remain so here so from here to here for this part he traveled with a constant velocity, constant velocity, v equal to 182 .59 miles per hour.
02:48
So he traveled with the same velocity for the remaining journey.
02:52
See again once i am repeating, initially he started with a zero velocity and he accelerated at this rate.
03:01
He accelerated with an acceleration rate a and he reaches a final velocity 182 .59 at some point and after that he traveled with the same 182 .59 velocity for the remaining journey.
03:16
So the total journey can be divided into two parts.
03:19
One part is from a to c so this is part 1 and from part c to b this is part 2.
03:27
So part 2 is consistent velocity.
03:31
Part 2 is constant velocity and part 1 is uniform velocity because the velocity is changing uniformly so this is part one is uniform velocity part and part two is constant velocity suppose consider so suppose this distance is now i am representing the distance suppose from part to a to c the distance is s cac and from c to b the distance is s c b from c to b this is the distance and time taken from point a to point c to point c to reach from point a to point c is taca and time taken from point to reach from point c to point b is t cb.
04:21
So i have completely represented the above question in this nice diagram.
04:26
You can understand it clearly.
04:28
Now first let us try to find the acceleration.
04:31
Now i am doing with this underpin to find, i am doing it here to find acceleration.
04:45
A equal to change in velocity by change in time.
04:53
We have given that change in velocity.
04:55
See, initially it started at rest and reaches 60 miles per hour.
04:59
So 60 minus 0.
05:02
And the time is 0, 4 .2.
05:05
So delta t is 4 .2 seconds.
05:10
And one more thing.
05:11
Since the time here, it was given in seconds, but here the velocity is in 60 mile per hour, we should convert this time into hour.
05:20
So 4 .2 seconds i am converting 4 .2.
05:24
3 ,600 hour because one second equal we have the conversion i am writing it here conversion one hour equal to 3 ,600 seconds therefore one second equal to 1 by 3 ,600 hour so substituted 1 by 3 ,600 in place of 1 second so it becomes like this so if you do this one you get the final answer as 51428 .57 miles per hour square.
06:07
So this is the acceleration in miles per hour square.
06:10
So this is the acceleration.
06:17
Now we have to find the time for both parts and if we add total, both parts tic and t -c we get total time.
06:26
Now for doing that, let us use our basic concept of physics.
06:34
So i am doing that.
06:35
Removing conversion here, please note it down because, okay, so i'm removing.
06:46
Okay, now, for part ac, for part ac, since part ac is uniform velocity part, we have the equations of motion.
06:58
We have the equations of motion.
07:04
We have equations of motion.
07:08
Three equations.
07:09
One is v equal to u plus a .t.
07:13
This is the first and v square minus u square equal to 2a s this is second one and s equal to u t plus half a t square this is the third one so we have three equations of motion so from this for part two first we should use you can use any of the three equations but we want distance and time two to find time you can use equation 1 and to find distance you can use equation 3.
07:52
So i am writing here from 1, from equation 1, v is the final velocity.
08:00
It is 182 .59 miles per hour equal to initial velocity u is 0 plus a acceleration value...