The acceleration of a particle as it moves along a straight...

The acceleration of a particle as it moves along a straight line is given by $a=(2 t-1) \mathrm{m} / \mathrm{s}^{2},$ where $t$ is in seconds. If $s=1 \mathrm{m}$ and $v=2 \mathrm{m} / \mathrm{s}$ when $t=0,$ determine the particle's velocity and position when $t=6 \mathrm{s}$. Also, determine the total distance the particle travels during this time period.

Key Concepts

Application of Initial Conditions

Initial conditions provide the specific values for velocity and position at a given starting time, usually necessary to determine the constants of integration when solving differential equations. Applying these conditions customizes the general solution to the specific motion scenario described.

Displacement versus Total Distance Traveled

Understanding the difference between displacement (the net change in position) and the total distance traveled (the sum of all distances moved, regardless of direction) is crucial. This concept is particularly important in cases where a moving particle reverses its direction, necessitating an absolute value integration to compute the actual distance covered.

Integration of Differential Equations

Integration is used to reverse the process of differentiation. In kinematics, integrating the acceleration function with respect to time gives the velocity function, and integrating the velocity function gives the position function. This process is central to solving problems where the rate of change of a quantity is known.

Acceleration as a Function of Time

Acceleration, defined as the rate of change of velocity, can vary with time. When given as a function of time, the acceleration requires integration to determine the velocity function. This concept is essential in understanding how forces acting on a particle influence its motion over time.

Transcript

00:01 Well, the expression for instantaneous axolation is given by instantaneous ax relation is equal to dv divided by d t.

00:13 And from here, we can express dv is equal to x relation times d t.

00:24 Now we replace x relation by 2t minus 1.

00:27 Therefore we have dv is equal to 2 t minus 1 into d t now let's integrate both sides of equation integral of dv is equal to integral of 2 t minus 1 into d t well let's write down the limits the limits of velocity are from v0 to v.

01:02 The limits of time would be from t0 to t.

01:08 Right? so solving this integral, we have v minus v0 is equal to t squared minus t minus t0 square minus t0 square minus t -0 well we know that v0 is equal to 2 and t -not is equal to 0 therefore we have v minus 2 is equal to t square minus t minus 0 square minus 0 and from here v is equal to t square minus t plus two well let's call this equation number one right now so uh velocity at time t is equal to six seconds so velocity when time t is equal to six is equal to six is 2 which is equal to 32 meter per second.

02:33 Well, in order to determine the total distance traveled, we must check how many times the particle has changed its direction, how many times its speed was equal to 0.

02:52 Well, to do so, we will just replace v .0 in expression number 1.

02:58 So 0 is equal to t squared minus t plus two and now so t now it's a quadratic equation so t is equal to 1 plus minus root 1 square minus 4 multiply by 2 multiply by 1 right divided by 2 to multiply by 1 right divided by 2 to multiply by 1.

03:39 It's here it's 1 by 2 in fact...

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