So the area under a velocity time graph is distance or displacement, if you have done calculus yet you will understand that if you take the integral of a velocity function then you end up with displacement. Thats for later understanding however.
So this appears to be a right triangle so we can find the area of a triangle as:
0.5bh = A
Since our area is 10 meters lets alter our formula a bit to fit the situation:
Our base here is time and our height is velocity so:
0.5tv = Δx
So we can read off the graph that our velocity at the end, or our final velocity appears to be near 2.0 m/s
So we have v, and Δx so lets isolate for time by dividing by v and 0.5
t = Δx / 0.5v
Now lets plug all that in:
t = 10 / 0.5(2)
t = 10 seconds
Hope this helped!
I am pretty sure that the only statement which is true for particles of the medium of an earthquake P-wave is being shown in the option : b)vibrate parallel to the wave, forming compressions and rarefactions. As you know, it can be formed in two ways : from alternating compressions and rarefactions or primary wave. I bet you will agree with me.
Answer:
The induced current in the resistor is I = BLv/R
Explanation:
The induced emf ε in the long bar of length, L in a magnetic field of strength, B moving with a velocity, v is given by
ε = BLv.
Now, the current I in the resistor is given by
I = ε/R where ε = induced emf in circuit and R = resistance of resistor.
So, the current I = ε/R.
substituting the value of ε the induced emf, we have
I = ε/R
I = BLv/R
So, the induced current through the resistor is given by I = BLv/R
The particle has acceleration vector
We're told that it starts off at the origin, so that its position vector at 
is
and that it has an initial velocity of 12 m/s in the positive 
direction, or equivalently its initial velocity vector is
To find the velocity vector for the particle at time 
, we integrate the acceleration vector:
![\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cleft%5B12%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cjmath)
![\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20v%3D%5Cleft%5B12%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5C%2C%5Cvec%5Cjmath)
Then we integrate this to find the position vector at time :
![\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%2812%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%2B%5Cleft%28-2.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5Cright%29%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%5B%5Cdisplaystyle%5Cint_0%5Et%5Cleft%284.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5C%2C%5Cmathrm%20d%5Ctau%5Cright%5D%5C%2C%5Cvec%5Cjmath)
![\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath](https://tex.z-dn.net/?f=%5Cvec%20r%3D%5Cleft%5B%5Cleft%2812%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29t%2B%5Cleft%28-1.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5Cright%5D%5C%2C%5Cvec%5Cimath%2B%5Cleft%282.0%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%5C%2C%5Cvec%5Cjmath)
Solve for the time when the 
coordinate is 18 m:
At this point, the coordinate is
so the answer is C.
