Particle acceleration If a particle moves along a coordinate line with a constant acceleration a

step 1 So in this problem, we have following condition. So initially, a particle is moving at an initia

Particle acceleration If a particle moves along a coordinate...

Key Concepts

Constant Acceleration

This refers to a scenario in kinematics where the acceleration of a particle remains constant over time. It simplifies the analysis of motion since the change in the velocity is linear, and the displacement changes following a quadratic function. Understanding constant acceleration is crucial in solving problems involving uniformly accelerated motion.

Kinematic Equations

These are the equations that describe an object's motion under constant acceleration. They relate displacement, initial velocity, acceleration, and time. A typical form is s(t) = (1/2)at² + v?t + s?, which encapsulates the contribution of acceleration and initial conditions to the particle's position at any given time.

Quadratic Function

In the context of constant acceleration, the displacement as a function of time is a quadratic function. This means that its graph is a parabola, and the coefficients in the quadratic equation represent physical quantities like acceleration, initial velocity, and initial position. Recognizing the quadratic nature is key for designing strategies to extract these parameters from given data points.

Systems of Equations

When specific values of displacement at different times are provided, substituting these into the kinematic equation produces a system of linear equations in the unknowns (acceleration, initial velocity, and initial position). Solving this system systematically through methods like substitution or elimination allows one to determine the unknown parameters in the equation of motion.

Transcript

00:02 So in this problem, we have following condition.

00:07 So initially, a particle is moving at an initial velocity of p .o and started at a position s of o from the origin and moving at the constant acceleration a.

00:21 Now, take note that if a particle is moving at a constant acceleration, then its distance with respect the time is being obeyed by this equation right here.

00:33 S of t is equal to one half a t squared plus b .o t plus s of o, wherein, as i've mentioned, s of o is the initial position, b of o is the initial velocity, and a is the constant acceleration.

00:47 If the acceleration is not constant, then this equation cannot be used.

00:52 Now, the end goal for this problem is to identify what the values of a, b, of o, and s of o are.

00:58 If after one half second or s of one half the distance traveled is seven if after one second or s of one second or s of one the distance traveled is 11 and after t is equal three halves seconds the distance travel is 17 meters the units are not mentioned here but just using the assumptions that this equation uses the s i unit so i'm assuming time is in second and distance is in meter.

01:33 Okay, so what we're going to do here is that we substitute each of the following condition.

01:38 So at time is equal to one half, s of t is equal to seven.

01:46 Using the equation we have seven is equal to one half a time given is one half, square that, these of all times one half.

01:58 I'm just plugging in the values of t and s of t in the equation.

02:02 So here we'll say 7 is equal to 1 over 8a plus 1 half piece of 0 plus s of all.

02:15 And let me say s of o is equal to 1 over 8, sorry, s of o is equal to 7 minus 1 over 8a minus 1 half piece of o.

02:32 And let me call it equation number 1.

02:38 The time is equal to 1, s of t is 7.

02:43 Sorry, is 11.

02:45 So we have 11 here for s of t.

02:49 And then plugging in the time on the equation, we have 1 half a times 1 squared plus b of o times 1 plus s of o.

02:58 So we'll end up having 11 is equal to 1 half a plus b of o.

03:05 And let me call it equation number 2.

03:09 Now when time is equal to three halves, s of d is 17.

03:16 So again, this value, so the equation we have 17.

03:20 It's equal to one half a times the time of three halves, square that, plus b's 0 times three halves plus s of o.

03:32 17 here is equal to nine over eight.

03:38 So three halves squared is nine over four times one half, so that should be nine over eight, a, plus three halves b s of o plus s of o.

03:51 And let me call it equation number three.

03:56 Okay.

03:56 So what i wanted to do now is the substitute equation number one to equations two and three so that i can minimize the number of variables.

04:08 So substitute one, equation number two...

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