Solve this following
Question: Prove that $\left|\begin{array}{ccc}a b-c c+b \\ a+c b c-a \\ a-b b+a c\end{array}\right|=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$ Solution:...
Read More →A bullet travelling with a velocity of
Question: A bullet travelling with a velocity of $16 \mathrm{~m} / \mathrm{s}$ penetrates a tree trunk and comes to rest in $0.4 \mathrm{~m}$. Find the time taken during the retardation. Solution: $\mathrm{u}=16 \mathrm{~m} / \mathrm{s} ; \mathrm{v}=0 \mathrm{~m} / 2 ; \mathrm{s}=0.4 \mathrm{~m}$ $v^{2}=u^{2}+2 a s$ $0^{2}=(1)^{2}+2(a)(0.4)$ $a=-320 \mathrm{~m} / \mathrm{s}^{2}$ $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $0=16+(-320) t$ $t=0.05 \mathrm{sec}$...
Read More →A train starts from rest and moved with a constant acceleration of
Question: A train starts from rest and moved with a constant acceleration of $2.0 \mathrm{~m} / \mathrm{s}^{2}$ for half a minute. The brakes are then applied and the train comes to rest in one minute. (a) Find the total distance moved by the train (b) The maximum speed attained by the train (c) The position(s) of the train at half the maximum speed. Solution: $A=$ slope $=\frac{v}{t}$ $2=\frac{v}{30}$ $\mathrm{v}=60 \mathrm{~m} / \mathrm{s}$ (a) Distance $=$ Area of $v-t$ graph $=\frac{1}{2}(30...
Read More →Solve the following equations:
Question: Solve the following equations: $\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -3 2 x x+2\end{array}\right|=0$ Solution: $\therefore x=1$ or $x=-3$ or $x=2$...
Read More →Solve the following equations:
Question: Solve the following equations: $\left|\begin{array}{ccc}x 3 7 \\ 2 x 2 \\ 7 6 x\end{array}\right|=0$ Solution:...
Read More →A person travelling at
Question: A person travelling at $43.2 \mathrm{~km} / \mathrm{h}$ applies the brake giving a deceleration of $6.0 \mathrm{~m} / \mathrm{s}^{2}$ to his scooter. How far will it travel before stopping? Solution: $\mathrm{u}=43.2^{\times \frac{5}{18}}=12 \mathrm{~m} / \mathrm{s} ; \mathrm{v}=0 ; \mathrm{a}=-6 \mathrm{~m} / \mathrm{s}^{2}$ Using, $v^{2}=u^{2}+2 a s$ $\mathrm{O}^{2}=(12)^{2}-2^{\mathrm{X}_{6}} \mathrm{X}_{\mathrm{s}}$ $s=12 m$...
Read More →An object having a velocity
Question: An object having a velocity $4.0 \mathrm{~m} / \mathrm{s}$ is accelerated at the rate of $1.2 \mathrm{~m} / \mathrm{s}^{2}$ for $5.0 \mathrm{~s}$. Find the distance travelled during the period of acceleration. Solution: $\mathrm{u}=4 \mathrm{~m} / \mathrm{s} ; \mathrm{a}=1.2 \mathrm{~m} / \mathrm{s}^{2} ; \mathrm{t}=5 \mathrm{sec}$ Distance travelled $\mathrm{s}=u t^{\frac{1}{2}} a t^{2}$ $s=(4)(5)+\frac{\frac{1}{2}}{2}(1.2)(5)^{2}$ $=35 \mathrm{~m}$...
Read More →A particle starts from a point A and travels along
Question: A particle starts from a point $A$ and travels along the solid curve shown in the given figure. Find approximately the position $\mathrm{B}$ of the particle such that the average velocity between the positions $\mathrm{A}$ and $\mathrm{B}$ has the same direction as the instantaneous velocity at $\mathrm{B}$. Solution: Direction of instantaneous velocity of point $\mathrm{B}$ must be same as direction of average velocity $\overrightarrow{\mathrm{AB}}$. So, point is approximately $(5,3)$...
Read More →The given figure shows x-t graph of a particle.
Question: The given figure shows $x-t$ graph of a particle. Find the time $t$ such that the average velocity of the particle during the period 0 to $\mathrm{t}$ is zero. Solution: Average velocity is zero when displacement is zero At $t=0 ; x=20$ and again at $t=12 ; x=20$...
Read More →Solve the following equations:
Question: Solve the following equations: $\left|\begin{array}{ccc}x+1 3 5 \\ 2 x+2 5 \\ 2 3 x+4\end{array}\right|=0$ Solution:...
Read More →From the velocity-time plot shown in the given figure,
Question: From the velocity-time plot shown in the given figure, find the distance travelled by the particle during the first 40seconds. Also find the average velocity during the period. Solution: Distance $=$ Area of $v$-t graph $=\left(\frac{1}{2} \times 20 \times 5\right)+\left(\frac{1}{2} \times 20 \times 5\right)$ $=100 \mathrm{~m}$ Displacement $=\left(\frac{1}{2} \times 20 \times 5\right)+\left(-\frac{1}{2} \times 20 \times 5\right)$ Displacement $=0$ $\overrightarrow{V_{\text {ang }}}=\f...
Read More →The given figure shows the graph of x-coordinate of
Question: The given figure shows the graph of x-coordinate of a particle going along the X-axis as a function of time. Find (a) The average velocity during 0 to 10s (b) Instantaneous velocity at 2, 5, 8 and 12s. Solution: (a) Velocity=displacement/time $=\frac{100}{10}=10 \mathrm{~m} / \mathrm{s}$ (b) Instantaneous velocity=slope of v-t graph At $t=2.5 \mathrm{~s} ;$ slope $=\frac{50-0}{2.5-0}=20 \mathrm{~m} / \mathrm{s}$ At $t=5 \mathrm{~s} ;$ slope $=0 \mathrm{~m} / \mathrm{s}$ At t $=8 \mathr...
Read More →Solve the following equations:
Question: Solve the following equations: $\left|\begin{array}{ccc}3 x-8 3 3 \\ 3 3 x-8 3 \\ 3 3 3 x-8\end{array}\right|=0$ Solution: $\therefore x=\frac{2}{3}$ or $x=\frac{11}{3}$...
Read More →The given figure shows the graph of velocity versus time
Question: The given figure shows the graph of velocity versus time for a particle going along the X-axis. Find (a) The acceleration (b) The distance travelled in 0 to 10 s and (c) The displacement in 0 to $10 \mathrm{~s}$ Solution: (a) Acceleration=slope of $v-t$ graph $=\frac{(8-2)}{10}=0.6 \mathrm{~m} / \mathrm{s}^{2}$ (b) Distance travelled = Area under v-t graph $=\frac{1}{2}(2+8) 10$ $=50 m$ (c) Displacement $=50 \mathrm{~m}$...
Read More →The acceleration of a cart started at t=0,
Question: The acceleration of a cart started at t=0, varies with time as shown in the given figure. Find the distance travelled in 30 seconds and draw the position-time graph. Solution: By velocity-time graph, Acceleration $=$ slope $=t$ $5=\frac{v}{10}$ $\mathrm{V}=50 \mathrm{~m} / \mathrm{s}$ Distance=Area of $v$-t graph $=\frac{1}{2}(30+10)(50)$ $=1000 \mathrm{ft}$...
Read More →The speed of a car as a function of time is shown in the given figure.
Question: The speed of a car as a function of time is shown in the given figure. Find the distance travelled by the car in 8 seconds and its acceleration. Solution: Acceleration=slope of $(v-t)$ graph $a=\tan \theta=\frac{20}{8}$ $=2.5 \mathrm{~m} / \mathrm{s}^{2}$ Distance=Area under (v-t) graph $=\frac{1}{2} \times 8 \times 20$ $=80 m$...
Read More →An athlete takes 2.0s to reach his maximum speed of 18.0 km/h.
Question: An athlete takes 2.0s to reach his maximum speed of 18.0 km/h. What is the magnitude of his average acceleration? Solution: $u=0 ; v=18 \times \frac{5}{18}=5 \mathrm{~m} / \mathrm{s} ; \mathrm{t}=2 \mathrm{sec}$ $\mathrm{v}=\mathrm{u}+\mathrm{at}$ $5=0+a(2)$ $\mathrm{a}=2.5 \mathrm{~m} / \mathrm{s}^{2}$...
Read More →When a person leaves his home for sightseeing by his car,
Question: When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hours the reading is 12416 km. (a) What is the average speed of the car during the period? (b) What is the average velocity? Solution: (a) $\mathrm{V}_{\mathrm{avg}}=\frac{\text { Total distance }}{\text { time }}=\frac{(12416-12352)}{2}$ $V_{\text {avq }}=32 \mathrm{kmph}$ (b) $\vec{V}$ =displacement $/$ time $\vec{V}_{\text {avg }}=0$...
Read More →Solve the following equations:
Question: Solve the following equations: $\left|\begin{array}{ccc}x+a b c \\ a x+b c \\ b b x+c\end{array}\right|=0$ Solution:...
Read More →Solve the following equations:
Question: Solve the following equations: $\left|\begin{array}{lll}1 x x^{3} \\ 1 b b^{3} \\ 1 c x^{3}\end{array}\right|=0$ Solution:...
Read More →Solve this following
Question: Show that $x=2$ is a root of the equation $\left|\begin{array}{ccc}x -6 -1 \\ 2 -3 x x-3 \\ -3 2 x 2+x\end{array}\right|=0$ Solution: $\rightarrow x=2$...
Read More →It is 260 km from Patna to Ranchi by air and 320 km by road.
Question: It is 260 km from Patna to Ranchi by air and 320 km by road. An aero plane takes 30 minutes to go from Patna to Ranchi whereas a deluxe bus takes 8 hours (a) Find the average speed of the plane. (b) Find the average speed of the bus. (c) Find the average velocity of the plane. (d)Find the average velocity of the bus. Solution: (a) $\vec{V}_{\text {avgpplane }}=\frac{\text { distance }}{\text { time }}=\frac{260}{0.5}=520 \mathrm{kmph}$ (b) $\vec{V}_{\text {avglius }}=\frac{\text { dist...
Read More →A particle starts from the origin goes along the X-axis
Question: A particle starts from the origin goes along the X-axis to the point (20m, 0) and th32en returns along the same line to the point (-20m, 0). Find the distance and displacement of the particle during the trip. Solution: Distance travelled $=20+40=60 \mathrm{~m}$ Displacement $=20 \mathrm{~m}$ in the negative direction...
Read More →A man has to go 50m due north,
Question: A man has to go 50m due north, 40m due east and 20m due south to reach a field. (a) What distance he has to walk to reach the field? (b) What is his displacement from his house to the field? Solution: (a) Distance travelled = 50+40+20 =110m (b) Displacement = 50 ĵ +40 - 20ĵ = 40+30 ĵ $|\operatorname{disp}|=\sqrt{(40)^{2}+(30)^{2}}$ $=50 \mathrm{~m}$ at an angle of $\Theta=\tan ^{-1}\left(\frac{30}{40}\right)=37^{\circ}$ north to east...
Read More →Solve the following :
Question: Find the center of mass of a uniform plate having semicircular inner and outer boundaries of radii $R_{1}$ and $R_{2}$. Solution: C.O.M of shaded region $=\frac{\int r \cdot d m_{R_{2}}-\int r . d m_{R_{1}}}{M_{R_{2}}-M_{R_{1}}}$ $M_{R_{\Omega}}=\int d m R_{2}=\rho\left(\frac{4 \pi R_{2}^{2}}{2}\right) l$ $M_{R_{1}}=\int d m R_{1}=\rho\left(\frac{4 \pi R_{1}^{2}}{2}\right) l$ ${ }_{\{} P=$ density of material $\}$ $\{\mid=$ thickness of disks $\}$ $d m_{R_{2}}=\rho l d A$ $=\rho \cdot ...
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