JEE Physics Motion In A Straight Line Notes Set 05

Revision Notes for JEE Physics: Motion In A Straight Line

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Reading Motion Graphs

GraphSlope givesArea gives
Position-time (x-t)VelocityNot used
Velocity-time (v-t)AccelerationDisplacement (signed); distance is the sum of absolute areas
Acceleration-time (a-t)Rate of change of accelerationChange in velocity
Velocity-position (v-x)\(\dfrac{dv}{dx}\), so \(a=v\dfrac{dv}{dx}\)Not used
\(v^2\)-position\(2a\)Not used

1. Zero slope on a smooth x-t curve means the particle is momentarily at rest; a sharp corner means a sudden change in velocity.
2. A v-t graph crossing the time axis means the direction of motion reverses at that instant.
3. A valid graph gives one value of the quantity for each value of time, so closed loops and curves that turn back in time are impossible.
4. Velocity and acceleration are continuous in practice, so sudden vertical jumps in a v-t graph are idealisations.

Distance, Displacement and Average Speed from Graphs

1. Split the v-t graph into triangles, rectangles and trapezia and add areas with signs for displacement.
2. Average speed is total distance over total time; average velocity is net displacement over total time.
3. If the time interval ends part-way through a motion, include only the part inside the interval.

Relative Motion in One Dimension

1. \(v_{AB}=v_A-v_B\) with signs. Opposite directions add the speeds, the same direction subtracts them.
2. A train of length L crossing a pole needs a distance L; crossing another train or a bridge needs the sum of the lengths.
3. Overtaking: write \(x_1(t)=x_2(t)\) from a common origin and clock; two roots mean they meet twice.
4. Two freely falling bodies have zero relative acceleration, so they approach each other at a constant speed equal to the difference of their initial velocities.
5. Escalator or walkway: speeds add, so \(\dfrac1t=\dfrac1{t_1}+\dfrac1{t_2}\).

Variable Acceleration

1. Given x(t): differentiate for v and a. The particle is at rest where v = 0, and its speed is maximum or minimum where a = 0.
2. Given a(t): integrate with the initial conditions. Given a(x): use \(v\,dv=a\,dx\).
3. If \(s\propto\sqrt t\) then \(v\propto t^{-1/2}\) and \(a\propto t^{-3/2}\propto v^3\), so differentiate and compare the powers of t.
4. To find the distance travelled (not displacement) find the turning points first and add the separate distances.

Free Fall Extras

1. Dropped from a moving carrier, the object starts with the carrier's velocity at that instant.
2. In a lift accelerating upwards with a, a falling object has relative acceleration \(g+a\) with respect to the lift.
3. With constant air resistance, the ascent time is smaller than the descent time (deceleration up is larger than down).
4. For two stones in succession: if the second is thrown after \(\Delta t\), write both position equations with the same origin.

Study Material and Revision Notes for JEE Physics Motion In A Straight Line

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