The velocity-time graph of a particle in 1 D motion is given...

The velocity-time graph of a particle in 1 D motion is given below. (i) Draw the acceleration-time graph from $\mathrm{t}=0$ to $\mathrm{t}=12 \mathrm{~s}$. (ii) Find displacement in the interval $\mathrm{t}=3 \mathrm{~s}$ to $\mathrm{t}=6 \mathrm{~s}$. (iii) Find displacement from $\mathrm{t}=0$ till $\mathrm{t}=12 \mathrm{~s}$. (iv) Find distance covered from $\mathrm{t}=0$ till $\mathrm{t}=12 \mathrm{~s}$.

The velocity-time graph of a particle in 1 D motion is given below.
(i) Draw the acceleration-time graph from $\mathrm{t}=0$ to $\mathrm{t}=12 \mathrm{~s}$.
(ii) Find displacement in the interval $\mathrm{t}=3 \mathrm{~s}$ to $\mathrm{t}=6 \mathrm{~s}$.
(iii) Find displacement from $\mathrm{t}=0$ till $\mathrm{t}=12 \mathrm{~s}$.
(iv) Find distance covered from $\mathrm{t}=0$ till $\mathrm{t}=12 \mathrm{~s}$.

Key Concepts

Velocity-Time Graph Interpretation

A velocity-time graph is a visual representation that illustrates the velocity of a particle as a function of time. It provides important information about the particle's motion by showing periods of constant velocity, acceleration, or deceleration. The slopes and areas under the curve are crucial for analyzing the particle’s behavior over time.

Acceleration-Time Graph Construction

Acceleration is defined as the rate of change of velocity with respect to time. To construct an acceleration-time graph from a velocity-time graph, one analyzes the slopes of the velocity-time graph at different time intervals. Constant slopes indicate constant acceleration, while changes in slope indicate varying acceleration, including transitions between acceleration and deceleration.

Displacement Calculation

Displacement is a vector quantity that represents the net change in position of a particle over a certain time interval. It can be determined by calculating the area under the velocity-time graph, taking into account the sign of the velocity. Positive areas correspond to displacement in one direction, and negative areas indicate displacement in the opposite direction.

Distance Covered Calculation

Distance covered is a scalar quantity that measures the total length of the path traveled by the particle, regardless of the direction of motion. It is calculated by taking the absolute value of the velocity at each moment before integrating over time. This process involves summing the absolute areas under the velocity-time graph over the specified time interval.

Transcript

00:01 In the given problem, first of all we are told to find out the graph between acceleration and time.

00:10 Exhaeration is basically dv upon d t which is slope of v versus t graph.

00:25 So to find out a t graph, first of all let us find out slope of each segment.

00:33 Slope of this and this segment is clearly zero.

00:39 This and this segment is positive and this segment has negative slope.

00:46 Now for this part of graph, slope is equal to 10 theta which is height upon base and height is 4 and base here is 2.

01:04 So 10 theta will be equal to 4 upon 2 equals 2.

01:11 Now for this segment slope is also 10 theta equals height upon base, height is 3 and base is 2.

01:27 So slope here is 1 .5 and this is also positive.

01:32 Now for this segment, from t equals to 4 to 2 equals to 8, slope is negative.

01:44 Let's find out the slope only for this segment and this slope will be the loaf of the entire segment.

01:54 Let's consider this triangle.

01:59 10 theta equals height upon base equals 8 upon 3 and xeration will be minus 8 upon 3.

02:11 Because slope here is negative.

02:16 Now let's make a t graph.

02:19 Suppose this is a and this is t.

02:23 For 0 to 2 acceleration is 0 and 4 2 acceleration is 2.

02:30 And for 2 to 4 acceleration is 2.

02:35 Suppose this is 2 and 4.

02:40 8 acceleration is minus 8x3 which is little bit greater than minus 2 in negative side.

02:51 This is minus 2.

02:53 So this will be approximately minus 8 by 3.

02:59 And for t equals to 8 to 10, acceleration is again 0.

03:06 And for t equals to 10 to 12, acceleration is 1 .5.

03:11 So this is acceleration versus time graph.

03:14 The first part, first sub part of this.

03:17 Discussion has been solved.

03:19 Now the second sub part of the equation says that distance covered by object between time t equals to 3 to 6.

03:31 Now distance is area under the graph of v versus t which can be calculated from the integration of v with respect to d.

03:52 Now to calculate the area of the desired portion, let us break it in portions.

04:00 And then we will add area of each segment...

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