Train A can move with a uniform acceleration of 3 m s^-2 while speeding up and with a uniform

Train A can move with a uniform acceleration of 3 m s^-2...

Train A can move with a uniform acceleration of $3 \mathrm{~m} \mathrm{~s}^{-2}$ while speeding up and with a uniform deceleration of $5 \mathrm{~m}$ $\mathrm{s}^{-2}$ on applying the brakes. Train $\mathrm{B}$ can move with a uniform acceleration of $4 \mathrm{~m} \mathrm{~s}^{-2}$ while speeding up and with a uniform deceleration of $4 \mathrm{~m} \mathrm{~s}^{-2}$ on applying the brakes. If $\mathrm{t}_{\mathrm{A}}$ and $\mathrm{t}_{\mathrm{B}}$ are the minimum times taken by train $\mathrm{A}$ and train $B$ respectively to start from one station and stop at the next station while travelling on straight tracks, then (a) $\mathrm{t}_{\mathrm{A}}=\mathrm{t}_{\mathrm{B}}$ (b) $\mathrm{t}_{\mathrm{A}}>\mathrm{t}_{\mathrm{B}}$ (c) $\mathrm{t}_{\mathrm{A}}<\mathrm{t}_{\mathrm{B}}$ (d) $\frac{\mathrm{t}_{\mathrm{A}}}{\mathrm{t}_{\mathrm{B}}}$ depends on the distance between stations.

Train A can move with a uniform acceleration of $3 \mathrm{~m} \mathrm{~s}^{-2}$ while speeding up and with a uniform deceleration of $5 \mathrm{~m}$ $\mathrm{s}^{-2}$ on applying the brakes. Train $\mathrm{B}$ can move with a uniform acceleration of $4 \mathrm{~m} \mathrm{~s}^{-2}$ while speeding up and with a uniform deceleration of $4 \mathrm{~m} \mathrm{~s}^{-2}$ on applying the brakes. If $\mathrm{t}_{\mathrm{A}}$ and $\mathrm{t}_{\mathrm{B}}$ are the minimum times taken by train $\mathrm{A}$ and train $B$ respectively to start from one station and stop at the next station while travelling on straight tracks, then
(a) $\mathrm{t}_{\mathrm{A}}=\mathrm{t}_{\mathrm{B}}$
(b) $\mathrm{t}_{\mathrm{A}}>\mathrm{t}_{\mathrm{B}}$
(c) $\mathrm{t}_{\mathrm{A}}<\mathrm{t}_{\mathrm{B}}$
(d) $\frac{\mathrm{t}_{\mathrm{A}}}{\mathrm{t}_{\mathrm{B}}}$ depends on the distance between stations.

Key Concepts

Uniform Acceleration

Uniform acceleration refers to a constant rate of change of velocity over time. In problems involving motion, it means that the acceleration value does not change during the acceleration phase, which simplifies the use of kinematic equations to determine parameters like time, displacement, and final velocity.

Uniform Deceleration

Uniform deceleration is the process of slowing down at a constant rate, essentially a constant negative acceleration. It plays a critical role in determining the time and distance required to stop a moving object from a given speed, and it uses the same kinematic principles as uniform acceleration but with a deceleration value.

Kinematic Equations

The kinematic equations are fundamental relationships in classical mechanics that connect displacement, initial velocity, final velocity, acceleration, and time. They are essential tools for analyzing motions under uniform acceleration or deceleration, allowing one to calculate unknown quantities when others are known.

Optimization of Motion Time

Optimization of motion time involves determining the minimum time needed for a trip by adjusting the phases of acceleration and deceleration. This concept analyzes at what point to switch from accelerating to decelerating so that a vehicle starting and ending at rest covers the distance in the least possible time, taking into account the different acceleration and deceleration capabilities.

Transcript

00:01 Hello students in this question train a can move with uniform acceleration of a1 equals to 3 meter per second square while speeding up and uniform the acceleration a2 equals to 5 meter per second square on a prime break and train b have these acceleration a1 and a2 both are equal to 4 meter per second square okay now we have given that t a and t b are the minimum time take by the train a and b to start and then stop to another station, then we have to determine which time is greater.

00:41 Okay.

00:42 So from the distance covered by both of them or train a during these two motions, this can be written as 1 by 2, mudplied bay, v, mudplied bay total time taken, that is t1 plus t2.

00:58 T1 is the time taken during this motion and this is time taken suppose t2.

01:02 V is the maximum speed which is obtained.

01:08 So we can write here that 1 by 2 and from the equation of motion if train is starting from the rest and finally stopping, so this a1 multiplied by t1, this will be equals to v and this will be also equal to a2 multiplied by t2.

01:22 So we can write here that we will be equals to a1, t1, okay? so this v can be replaced in the term.

01:33 Of these two parameters.

01:35 So from here we can write that the time total.

01:39 So these times can be represent in the total time term.

01:43 So v.

01:45 Mcdipler by this can be written as t1 from here.

01:48 So v.

01:49 B divided by a1 plus v...