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Physics 17 Online
OpenStudy (abhisar):

@YanaSidlinskiy

OpenStudy (yanasidlinskiy):

29=33+a*2.5

OpenStudy (abhisar):

Ok, so I don't know if you have read the theoretical part. But I'll like to discuss few basic things first before we solve this question. Ok?

OpenStudy (abhisar):

Do you understand acceleration/deacceleration ?

OpenStudy (yanasidlinskiy):

Acceleration is when the rate inscreases and deacceleration is when spead decreases

OpenStudy (abhisar):

Yes, there are two types of motion. One in which the body keeps moving with a constant velocity. Like if its initial velocity is 33 m/s then it will keep moving with this same velocity. The other is accelerated motion in which its velocity will keep increasing at a certain rate.

OpenStudy (abhisar):

Cool?

OpenStudy (yanasidlinskiy):

Ok

OpenStudy (abhisar):

Now, while solving numerical problems. You just need to keep in mind these 3 equations. 1) V=U+at 2) \(\sf S=Ut + 1/2at^2\) 3) \(V^2=U^2+2as\) Where, V=final velocity U=Initial Velocity a=acceleration t=time

OpenStudy (abhisar):

and S=distance traveled.

OpenStudy (abhisar):

These are popularly called as Equations of motion. Cool?

OpenStudy (yanasidlinskiy):

Yes

OpenStudy (abhisar):

Ok, let's start solving the easy pisy question. Read your original question and note down what information is given first.

OpenStudy (abhisar):

Initial velocity=? Final velocity=? acceleration=? time=? distance=?

OpenStudy (yanasidlinskiy):

Can I go shower lol

OpenStudy (abhisar):

You can wait for 10 mins. That's what this would take.

OpenStudy (yanasidlinskiy):

Ok ok fine..

OpenStudy (yanasidlinskiy):

I'm sick of this problem.

OpenStudy (yanasidlinskiy):

IV= 33 FV = 29 A= NOT GIVEN Time=2.5 Distance= Idk

OpenStudy (abhisar):

Yes, that's right! You need to find the distance.

OpenStudy (abhisar):

Now see which equation from the above can you use to find S?

OpenStudy (yanasidlinskiy):

2nd

OpenStudy (abhisar):

There are two equations which has S in it.

OpenStudy (abhisar):

You can use either 2nd or 3rd. Right?

OpenStudy (yanasidlinskiy):

I think so..

OpenStudy (abhisar):

But see, you can't use then directly because both of them requires info about acceleration and we don't have it. Correct?

OpenStudy (yanasidlinskiy):

Yes

OpenStudy (abhisar):

So, we will have to use the 1st equation to find a and then substitute its value in either 2nd or 3rd equation to get the value of S

OpenStudy (abhisar):

But, I gave you an easy way. \(\sf v=u+at \\ (v-u)/t = a ...... (1)\) Substituting this value of a in 2nd equation of motion \(\sf S=ut+1/2 \frac{v-u}{t} \times t^2\\ S=\frac{v+u}2{} \times t\)

OpenStudy (abhisar):

Simply put the values in the last equation and you will get the answer.

OpenStudy (yanasidlinskiy):

WTF is U and

OpenStudy (abhisar):

U=initial velocity V=final velocity S=distance traveled t=time a=acceleration

OpenStudy (yanasidlinskiy):

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