A particle starts from origin at t=0 with a velocity 5 î m s^-1 and moves in x-y plane under t

A particle starts from origin at t=0 with a velocity 5 î m...

A particle starts from origin at $t=0$ with a velocity $5 \hat{i} \mathrm{~m} \mathrm{~s}^{-1}$ and moves in $x-y$ plane under the action of a force which produces a constant acceleration of $3 \hat{i}+2 \hat{j} \mathrm{~m} \mathrm{~s}^{-2}$. The $y$-coordinate of the particle at the instant when its $x$-coordinate is $84 \mathrm{~m}$ is (a) $12 \mathrm{~m}$ (b) $24 \mathrm{~m}$ (c) $36 \mathrm{~m}$ (d) $48 \mathrm{~m}$

A particle starts from origin at $t=0$ with a velocity $5 \hat{i} \mathrm{~m} \mathrm{~s}^{-1}$ and moves in $x-y$ plane under the action of a force which produces a constant acceleration of $3 \hat{i}+2 \hat{j} \mathrm{~m} \mathrm{~s}^{-2}$. The $y$-coordinate of the particle at the instant when its $x$-coordinate is $84 \mathrm{~m}$ is
(a) $12 \mathrm{~m}$
(b) $24 \mathrm{~m}$
(c) $36 \mathrm{~m}$
(d) $48 \mathrm{~m}$

Key Concepts

Uniformly Accelerated Motion

This concept refers to motion where an object experiences a constant acceleration. It is central to solving problems in kinematics because it allows the use of standard equations of motion that relate displacement, initial velocity, time, and acceleration in a predictable, linear manner.

Kinematic Equations

These are the mathematical formulas used to relate displacement, velocity, acceleration, and time under constant acceleration. They are crucial for solving problems by allowing us to determine one variable when others are known, whether it is projectile motion in one dimension or components in two dimensions.

Component-wise Analysis

In two-dimensional motion, the motion is analyzed separately in the x and y directions. Each component of the motion is treated independently using the kinematic equations, assuming that the forces acting in perpendicular directions do not interact, which simplifies the problem into one-dimensional analyses along each axis.

Time Determination from Displacement

This involves calculating the time at which a particular displacement is achieved in one direction. Once the time is known from the motion in one axis, it is used in the corresponding motion equation in the perpendicular axis to find the displacement or other parameters at that same time.

Transcript

00:01 It is given that a particle starts from origin with a velocity of 5 i cap and it moves in xy plane under the action of a force which produce a constant acceleration.

00:10 Now we have to find the y coordinate of the particle at the instant when its x coordinate is 84 meter.

00:19 So we can say that here we have given that the initial velocity of the particle in x -daction that is v0 vector is equal to this is 5 i -cap meter per second and we can say that here we have given that the initial velocity of the particle in x -daction that is v0 vector is equal to this is 5 i cap meter per second and we can we have given the acceleration that is a vector is equal to this is 3 i cap plus 2 j cap meter per second square now we know that the position vector of the particle that is r vector is equal to this is this can be written as that is v0 vector into t plus half of a vector into t square so we can say this will come out to be that is r vector is equal to this is 5 t this is 5 this is 5 i cap into t plus this is half of 3 i cap plus 2 j cap into t square or we can say this is equals to 5 t plus this is 1 .5 t square into i cap plus this is t square into j cap so this is the position vector now we have given that we know that the position vector is given as that is r vector is equal to this is x i cap plus y j cap so if we compare both the equation so we can say that on comparing both we get that x is equal to this is 5 t plus 1 .5 of t square and the value of y is equal to this is t square now we have given that the position is for 84 meters so we have given that x is 84 meter this is given so we can say that from here we get this is 84 is equal to this is 5 t plus 1 .5 t square.

02:33 On solving this we get, so this is 1 .5 t squared plus 5 t minus 84 is equal to this is 0...

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