What can you say about the motion of an object whose distance time graph is a straight line parallel to the time axis

Question 1. An object has moved through a distance. Can it have zero displacement ? If yes, support your answer with an example. (CBSE 2015) Answer: Yes, The displacement of the object can be zero. Let a boy completes one round of a circular track in 5 minutes. The distance travelled by the boy = circumference of the circular track. However, displacement of the boy is zero because his initial and final positions are same. Question 2. A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds ? (CBSE 2010, 2015) Answer: ABCD is a square field of side 10 m. The farmer moves along the boundary of the field from the corner A via the corners B, C and D. After every 40 s, the farmer is again at the corner A, so his displacement after every 40 s is zero. At the end of 2 minutes 20 seconds = (2 x 60 + 20) = 140 s, the farmer will be at the corner C. Question 3. Distinguish between speed and velocity. (CBSE 2010, 2012, 2013, 2015) Answer: Speed Velocity 1. Distance travelled by an object per unit time is known as its speed. The distance travelled by an object in a particular direction (i.e. displacement) per unit time is known as its velocity. 2. Average speed of a moving object cannot be zero. Average velocity of a moving object can be zero. 3. tells how fast an object moves. Velocity tells how fast an object moves and in which direction it moves. 4. Speed is a scalar quantit...

Many people feel about graphs the same way they do about going to the dentist: a vague sense of anxiety and a strong desire for the experience to be over with as quickly as possible. But position graphs can be beautiful, and they are an efficient way of visually representing a vast amount of information about the motion of an object in a conveniently small space. The slope of this position graph is slope = rise run = x 2 − x 1 t 2 − t 1 \text slope = run rise ​ = t 2 ​ − t 1 ​ x 2 ​ − x 1 ​ ​ start text, s, l, o, p, e, end text, equals, start fraction, start text, r, i, s, e, end text, divided by, start text, r, u, n, end text, end fraction, equals, start fraction, x, start subscript, 2, end subscript, minus, x, start subscript, 1, end subscript, divided by, t, start subscript, 2, end subscript, minus, t, start subscript, 1, end subscript, end fraction . This expression for slope is the same as the definition of velocity: v = Δ x Δ t = x 2 − x 1 t 2 − t 1 v=\dfrac v = Δ t Δ x ​ = t 2 ​ − t 1 ​ x 2 ​ − x 1 ​ ​ v, equals, start fraction, delta, x, divided by, delta, t, end fraction, equals, start fraction, x, start subscript, 2, end subscript, minus, x, start subscript, 1, end subscript, divided by, t, start subscript, 2, end subscript, minus, t, start subscript, 1, end subscript, end fraction . So the slope of a position graph has to equal the velocity. The slope of the curve between the times t = 0 s t=0\text t = 3 s t, equals, 3, start text, space, s, end text is positive...

The vertical axis represents the velocity of the object. This probably sounds obvious, but be forewarned—velocity graphs are notoriously difficult to interpret. People get so used to finding velocity by determining the slope—as would be done with a position graph—they forget that for velocity graphs the value of the vertical axis is giving the velocity. slope = rise run = v 2 − v 1 t 2 − t 1 = Δ v Δ t \text slope = run rise ​ = t 2 ​ − t 1 ​ v 2 ​ − v 1 ​ ​ = Δ t Δ v ​ start text, s, l, o, p, e, end text, equals, start fraction, start text, r, i, s, e, end text, divided by, start text, r, u, n, end text, end fraction, equals, start fraction, v, start subscript, 2, end subscript, minus, v, start subscript, 1, end subscript, divided by, t, start subscript, 2, end subscript, minus, t, start subscript, 1, end subscript, end fraction, equals, start fraction, delta, v, divided by, delta, t, end fraction This means that when the slope is steep, the object will be changing velocity rapidly. When the slope is shallow, the object will not be changing its velocity as rapidly. This also means that if the slope is negative—directed downwards—the acceleration will be negative, and if the slope is positive—directed upwards—the acceleration will be positive. Δ x = v Δ t = ( 6 m/s ) ( 5 s ) = 30 m \Delta x=v\Delta t=(6\text Δ x = v Δ t = ( 6 m/s ) ( 5 s ) = 3 0 m delta, x, equals, v, delta, t, equals, left parenthesis, 6, start text, space, m, slash, s, end text, right parenthesis, left pa...

Distance-time graphs of motion Distance-time graphs Distance-time graphs show how the distance travelled by a moving object changes with time. These graphs also show if the object is moving at a constant speed or accelerating, going back to the starting point, or stationary. Constant speed is shown by a straight rising line. The gradient or the steepness of the graph can be used to work out the speed. A steep line means the vehicle is moving fast. A line curving upwards means that the vehicle is accelerating. The vehicle is getting faster as the line is getting steeper. A flat horizontal line means the vehicle is still, as the distance isn’t changing. The speed of the vehicle following the green line for the first 3 seconds is speed = distance ÷ time = gradient = 6 m ÷ 3 s = 2 m/s. The total distance travelled by the vehicle on the green line is 7 m. The mean speed of the vehicle on the green line is, average speed = total distance ÷ total time = 7 m ÷ 10 s = 0.7 m/s. The speed of the vehicle following the purple line for the first 2 seconds is speed = distance ÷ time = gradient = 10 m ÷ 2 s = 5 m/s. Look at this distance-time graph and answer the following questions. Question How far did the vehicle travel in the first 4 seconds? Reveal answer