Solve\begin {align*} \frac {dy}{dx} & =x\sqrt {x^{2}+9}\\ y\left ( -4\right ) & =0 \end {align*}
Solution
This is separable ODE. Integrating both sides gives\begin {equation} y\left ( x\right ) =\int x\sqrt {x^{2}+9}dx+c \tag {1} \end {equation} Where \(c\) is constant of integration. To integrate \(\int x\sqrt {x^{2}+9}dx\), let \(u=x^{2}+9\). Hence \(\frac {du}{dx}=2x\) or \(dx=\frac {du}{2x}\). Therefore the integral becomes\begin {align*} \int x\sqrt {x^{2}+9}dx & =\int x\sqrt {u}\frac {du}{2x}\\ & =\frac {1}{2}\int u^{\frac {1}{2}}du\\ & =\frac {1}{2}\frac {u^{\frac {3}{2}}}{\frac {3}{2}}\\ & =\frac {1}{2}\frac {2}{3}u^{\frac {3}{2}}\\ & =\frac {1}{3}u^{\frac {3}{2}} \end {align*}
But \(u=x^{2}+9\), hence the above becomes\[ \int x\sqrt {x^{2}+9}dx=\frac {1}{3}\left ( x^{2}+9\right ) ^{\frac {3}{2}}\] Substituting the above in (1) gives\begin {equation} y\left ( x\right ) =\frac {1}{3}\left ( x^{2}+9\right ) ^{\frac {3}{2}}+c \tag {2} \end {equation} The constant \(c\) is from initial conditions. Since \(y\left ( -4\right ) =0\) then Eq (2) becomes\begin {align*} 0 & =\frac {1}{3}\left ( 16+9\right ) ^{\frac {3}{2}}+c\\ & =\frac {1}{3}\left ( 25\right ) ^{\frac {3}{2}}+c\\ & =\frac {1}{3}\left ( 5^{2}\right ) ^{\frac {3}{2}}+c\\ & =\frac {1}{3}\left ( 5\right ) ^{3}+c\\ & =\frac {125}{3}+c \end {align*}
Hence \(c=-\frac {125}{3}\). Therefore the solution (2) becomes\begin {align*} y\left ( x\right ) & =\frac {1}{3}\left ( x^{2}+9\right ) ^{\frac {3}{2}}-\frac {125}{3}\\ & =\frac {1}{3}\left ( \left ( x^{2}+9\right ) ^{\frac {3}{2}}-125\right ) \end {align*}
Solve\begin {align*} \frac {dy}{dx} & =\cos 2x\\ y\left ( 0\right ) & =1 \end {align*}
Solution
This is separable ODE. Integrating both sides gives\begin {align} y\left ( x\right ) & =\int \cos 2xdx+c\nonumber \\ & =\frac {1}{2}\sin \left ( 2x\right ) +c \tag {1} \end {align}
The constant \(c\) is from initial conditions. Since \(y\left ( 0\right ) =1\) then (1) becomes\begin {align*} 1 & =\frac {\sin \left ( 0\right ) }{2}+c\\ & =c \end {align*}
Hence the solution (1) becomes\[ y\left ( x\right ) =\frac {1}{2}\sin \left ( 2x\right ) +1 \]
A ball is dropped from the top of a building \(400\) ft high. How long does it take to reach the ground? With what speed does the ball strike the ground?
Solution
Let the ground by level \(0\) (i.e. \(y=0\)) and let up be positive and down negative. Therefore \(y\left ( 0\right ) =400\) ft and assuming initial velocity is zero then \(y^{\prime }\left ( 0\right ) =v\left ( 0\right ) =0\).. Therefore\[ v\left ( t\right ) =\int a\left ( t\right ) dt \] Where \(a\left ( t\right ) \) is the acceleration, which in this case is \(g=-32\) ft/sec\(^{\text {2}}\). The above becomes\begin {align} v\left ( t\right ) & =-32t+v\left ( 0\right ) \tag {1}\\ & =-32t\nonumber \end {align}
And \begin {align*} y\left ( t\right ) & =\int v\left ( t\right ) dt\\ & =\int -32t\ dt\\ & =-\frac {32}{2}t^{2}+y\left ( 0\right ) \end {align*}
But \(y\left ( 0\right ) =400\,\)ft. The above becomes\[ y\left ( t\right ) =-16t^{2}+400 \] To find the time it takes to hit the ground, the above is solved for \(y\left ( t\right ) =0\). This gives\begin {align*} 0 & =-16t^{2}+400\\ t^{2} & =\frac {400}{16}\\ & =25 \end {align*}
Therefore the time is \(t=5\) seconds. Now we know how long it takes to reach the ground, we can find the velocity when ball strike the ground from (1). Substituting \(t=5\) in (1) gives\begin {align*} v\left ( 5\right ) & =-32\left ( 5\right ) \\ & =-160\text { ft/sec} \end {align*}
So it strikes the ground with speed \(160\) ft/sec in the downwards (negative) direction.
A projectile is fired straight upward with an initial velocity of \(100\) m/s from the top of a building \(20\) m high and falls to the ground at the base of the building. Find (a) its maximum height above the ground (b) when it passes the top of the building (c) its total time in the air.
Solution
Let the ground be level \(0\,\). (i.e. \(y=0\)) and let up be positive and down negative. Therefore \(y\left ( 0\right ) =20\) m. Initial velocity is \(100\) m/s, hence \(y^{\prime }\left ( 0\right ) =v\left ( 0\right ) =100\). The acceleration due to gravity is \(g=-9.8\) m/s\(^{\text {2}}\).\begin {align*} v\left ( t\right ) & =\int a\left ( t\right ) dt\\ & =-gt+v\left ( 0\right ) \\ & =-gt+100 \end {align*}
When the ball reaches maximum high above the building, it must have zero velocity. From the above this means\begin {align*} 0 & =-9.8t+100\\ t & =\frac {100}{g}\text { sec} \end {align*}
The above is how long it takes for the ball to reach maximum high. Now\begin {align*} y\left ( t\right ) & =\int v\left ( t\right ) dt\\ & =\int \left ( -gt+100\right ) dt+y\left ( 0\right ) \\ y\left ( t\right ) & =-\frac {1}{2}gt^{2}+100t+y\left ( 0\right ) \end {align*}
But \(y\left ( 0\right ) =20\). Therefore\[ y\left ( t\right ) =-\frac {1}{2}gt^{2}+100t+20 \] Substituting \(t=\frac {100}{g}\) in the above, gives the distance traveled above the ground until the ball reached maximum high. Therefore\begin {align*} y\left ( \frac {100}{g}\right ) & =-\frac {1}{2}g\left ( \frac {100}{g}\right ) ^{2}+100\left ( \frac {100}{g}\right ) +20\\ & =-\frac {1}{2}\frac {100^{2}}{g}+\frac {100^{2}}{g}+20\\ & =\frac {1}{2}\frac {100^{2}}{g}+20 \end {align*}
Using \(g=9.8\) the above gives\begin {align*} y\left ( \frac {100}{g}\right ) & =\frac {1}{2}\frac {100^{2}}{9.8}+20\\ y_{\max } & =530.2\text { meter} \end {align*}
The ball will take the same amount of time to fall down back to top of building, as the time it took to reach the maximum high above the building, since the distance is the same, and the acceleration is the same (gravity acceleration). This time is \(t_{0}=\frac {100}{g}\) sec found in part (a). Therefore, twice this time gives \begin {align*} t_{\text {travel}} & =\frac {200}{g}\\ & =\frac {200}{9.8}\\ & =20.408\text { sec} \end {align*}
Now we find the time it take to reach the ground. We now take initial velocity as \(v\left ( 0\right ) =0\), which is when the ball was at its maximum high above the building. And initial position is from part (a) was found \(y_{\max }=530.2\) meter. Hence \(y\left ( 0\right ) =530.2\) m. Now we will find the time to reach the ground, starting from the maximum high.\begin {align*} v\left ( t\right ) & =\int gdt\\ & =gt+v\left ( 0\right ) \\ & =gt \end {align*}
And \begin {align*} y\left ( t\right ) & =\int v\left ( t\right ) dt\\ & =\int gt\ dt+y\left ( 0\right ) \\ & =\frac {1}{2}gt^{2}+530.2 \end {align*}
When it hits the ground \(y\left ( t\right ) =0\),. Hence we now have an equation to solve for time\[ 0=\frac {1}{2}gt^{2}+530.2 \] But \(g=-9.8\). The above becomes\begin {align*} 0 & =\frac {1}{2}\left ( -9.8\right ) t^{2}+530.2\\ t^{2} & =\frac {2\left ( 530.2\right ) }{9.8} \end {align*}
Hence \(t=\sqrt {\frac {2\left ( 530.2\right ) }{9.8}}=10.402\) sec. This is the time it takes to fall to the ground, starting from maximum high. Adding to this time, the time it took to reach maximum high from top of building, which is \(\frac {100}{g}\) sec as found from part (a), gives total time in air\begin {align*} t_{\text {total}} & =10.402+\frac {100}{9.8}\\ & =20.606\text { sec} \end {align*}
Solution
Solution
A racecar accelerates from stationary at a rate of \(14\) m/s\(^{\text {2}}\). How long does it take the car to reach its top speed of \(300\) km/h? How far does the car travel in that time?
Solution
Let \(x\left ( 0\right ) =0,v\left ( 0\right ) =0\) and \(a=14\) m/s\(^{\text {2}}\). \begin {align*} v\left ( t\right ) & =\int a\left ( t\right ) dt\\ & =\int 14dt\\ & =14t+v\left ( 0\right ) \\ & =14t \end {align*}
Since we want to find time to reach \(v_{\text {max}}=300\) km/h which in SI units is \(\frac {\left ( 300\right ) \left ( 1000\right ) }{\left ( 60\right ) \left ( 60\right ) }=\frac {250}{3}\) m/sec. Substituting this in the above gives\begin {align*} \frac {250}{3} & =14t_{\text {max}}\\ t_{\text {max}} & =\frac {250}{3\left ( 14\right ) }\\ & =\frac {125}{21}\\ & =5.95\text { seconds} \end {align*}
To find the distance traveled in this time, since\begin {align*} x\left ( t\right ) & =\int v\left ( t\right ) dt\\ & =\int 14tdt\\ & =\frac {14}{2}t^{2}+x\left ( 0\right ) \\ & =7t^{2} \end {align*}
When \(t=t_{\text {max}}\) the above gives\begin {align*} x\left ( t_{\max }\right ) & =7\left ( \frac {125}{21}\right ) ^{2}\\ & =248.02\text { meters} \end {align*}
The car is approaching a tight turn at \(300\) km/h. In order to safely make the corner, it must be traveling at \(80\) km/h when it enters the corner. The brakes on the car cause a deceleration of \(39\) m/s\(^{\text {2}}\). How far away from the corner must the driver begin braking to make the corner?
Solution
In SI units \(300\) km/h is \(\frac {\left ( 300\right ) \left ( 1000\right ) }{\left ( 60\right ) \left ( 60\right ) }=\allowbreak \frac {250}{3}\) m/s. And \(80\) km/h is \(\frac {80\left ( 1000\right ) }{\left ( 60\right ) \left ( 60\right ) }=\allowbreak \frac {200}{9}\) m/s. Therefore we have initial velocity \(v\left ( 0\right ) =\frac {250}{3}\) m/s and final velocity \(v_{f}\left ( t\right ) =\frac {200}{9}\) m/s and have acceleration of \(-39\ \)m/s\(^{\text {2}}\).
We first find the time it takes to go from \(v\left ( 0\right ) \) to \(v_{f}\left ( t\right ) \). Since \begin {align*} v\left ( t\right ) & =\int a\left ( t\right ) dt\\ & =\int -39dt\\ & =-39t+v\left ( 0\right ) \end {align*}
Therefore we have the equation\begin {align*} v_{f}\left ( t\right ) & =-39t+v\left ( 0\right ) \\ \frac {200}{9} & =-39t+\frac {250}{3}\\ 39t & =\frac {250}{3}-\frac {200}{9}\\ t_{f} & =\frac {550}{351}\\ & =1.567\ \text {sec} \end {align*}
This is the time needed to decelerate from \(300\) km/h to \(80\) km/h. Now we find the distance traveled during this time. Since\begin {align*} x\left ( t\right ) & =\int v\left ( t\right ) dt\\ & =\int -39t+v\left ( 0\right ) dt\\ & =\int -39t+\frac {250}{3}dt\\ & =-\frac {39}{2}t^{2}+\frac {250}{3}t+x\left ( 0\right ) \end {align*}
Let \(x\left ( 0\right ) =0\), by taking initial position as zero. Replacing \(t\) in the above with \(t_{f}\) found earlier gives\begin {align*} x\left ( t\right ) & =-\frac {39}{2}\left ( 1.567^{2}\right ) +\frac {250}{3}\left ( 1.567\right ) \\ & =82.7\text { meter} \end {align*}
Therefore the car needs to be \(82.7\) meter away from corner to begin the braking.
At the exit of the corner, two cars are traveling at \(100\) km/h, with car \(A\) being \(10\) m behind car \(B\). Out of the corner, car \(A\) accelerates at \(14\) m/s\(^{\text {2}}\) and car \(B\) accelerates at \(13\) m/s\(^{\text {2}}\). How much time does it take for car \(A\) to be right next to car \(B\)? How fast are the cars going when this happens? How far from the corner exit have they traveled?
Solution
Using SI units, \(100\) km/h is \(\frac {\left ( 100\right ) \left ( 1000\right ) }{\left ( 60\right ) \left ( 60\right ) }=\frac {250}{9}\) m/s. Let at \(t=0\), \(x_{A}\left ( 0\right ) =0\) and therefore \(x_{B}\left ( 0\right ) =10\), since car \(B\) is ahead by \(10\) meters initially. Let \(v_{A}\left ( 0\right ) =\frac {250}{9}\) m/s and also \(v_{B}\left ( 0\right ) =\frac {250}{9}\ \)m/s. We now need to determine the time, say \(t_{f}\), where \(x_{A}\left ( t_{f}\right ) =x_{B}\left ( t_{f}\right ) \). But for car \(A\) we have\begin {align*} v_{A}\left ( t\right ) & =\int a_{A}\left ( t\right ) dt\\ & =\int 14dt\\ & =14t+v_{A}\left ( 0\right ) \\ & =14t+\frac {250}{9} \end {align*}
And\begin {align} x_{A}\left ( t\right ) & =\int v_{A}\left ( t\right ) dt\nonumber \\ & =\int \left ( 14t+\frac {250}{9}\right ) dt\nonumber \\ & =\frac {14}{2}t^{2}+\frac {250}{9}t+x_{A}\left ( 0\right ) \nonumber \\ & =7t^{2}+\frac {250}{9}t \tag {1} \end {align}
Since \(x_{A}\left ( 0\right ) =0\). Now we do the same for car \(B\)\begin {align*} v_{B}\left ( t\right ) & =\int a_{B}\left ( t\right ) dt\\ & =\int 13dt\\ & =13t+v_{A}\left ( 0\right ) \\ & =13t+\frac {250}{9} \end {align*}
And\begin {align} x_{B}\left ( t\right ) & =\int v_{B}\left ( t\right ) dt\nonumber \\ & =\int \left ( 13t+\frac {250}{9}\right ) dt\nonumber \\ & =\frac {13}{2}t^{2}+\frac {250}{9}t+x_{B}\left ( 0\right ) \nonumber \\ & =\frac {13}{2}t^{2}+\frac {250}{9}t+10 \tag {2} \end {align}
Since \(x_{B}\left ( 0\right ) =10\) m. Now we solve for \(t\) by equating (1) and (2)\begin {align*} 7t^{2}+\allowbreak \frac {250}{9}t & =\frac {13}{2}t^{2}+\frac {250}{9}t+10\\ 7t^{2}+ & =\frac {13}{2}t^{2}+10\\ 7t^{2}-\frac {13}{2}t^{2} & =10\\ \frac {1}{2}t^{2} & =10\\ t^{2} & =20\\ t & =\sqrt {20}\\ t_{f} & =4.47\text { sec} \end {align*}
So it takes \(4.47\) sec for car \(A\) to be be next to car \(B\). To find the speed at this time, we substitute this value of time back in the velocity equation above. For car \(A\)\begin {align} v_{A}\left ( t\right ) & =14t+\frac {250}{9}\nonumber \\ v_{A}\left ( t_{f}\right ) & =14\left ( 4.47\right ) +\frac {250}{9}\nonumber \\ & =90.36\text { m/s} \tag {3} \end {align}
And for car \(B\)\begin {align*} v_{B}\left ( t\right ) & =13t+\frac {250}{9}\\ v_{B}\left ( t_{f}\right ) & =13\left ( 4.47\right ) +\frac {250}{9}\\ & =85.89\text { m/s} \end {align*}
To find the distance traveled during this time, we substitute this time in the position equation. For car \(A\), from Eq (1)\begin {align*} x_{A}\left ( t\right ) & =7t^{2}+\frac {250}{9}t\\ x_{A}\left ( t_{f}\right ) & =7\left ( 4.47\right ) ^{2}+\frac {250}{9}\left ( 4.47\right ) \\ & =264.03\text { meter} \end {align*}
The distance traveled by car \(B\) is \(10\) meters less than this value, since it was ahead by \(10\) meters at the start at time \(t=0\).