\nExample: mass, speed<\/td>\n | Example: force, velocity<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Solution 2S.<\/strong><\/span><\/p>\na) Pressure is a scalar quantity. \nb) Momentum is a vector quantity. \nc) Weight is a vector quantity. \nd) Force is a vector quantity. \ne) Energy is a scalar quantity. \nf) Speed is a scalar quantity.<\/p>\n Solution 3S.<\/strong><\/span><\/p>\nA body is said to be at rest if it does not change its position with respect to its immediate surroundings.<\/p>\n Solution 4S.<\/strong><\/span><\/p>\nA body is said to be in motion if it changes its position with respect to its immediate surroundings.<\/p>\n Solution 5S.<\/strong><\/span><\/p>\nWhen a body moves along a straight line path, its motion is said to be one-dimensional motion.<\/p>\n Solution 6S.<\/strong><\/span><\/p>\nThe shortest distance from the initial to the final position of the body is called the magnitude of displacement. It is in the direction from the initial position to the final position. \nIts SI unit is metre (m).<\/p>\n Solution 7S.<\/strong><\/span><\/p>\nDistance is a scalar quantity, while displacement is a vector quantity. The magnitude of displacement is either equal to or less than the distance. The distance is the length of path travelled by the body so it is always positive, but the displacement is the shortest length in direction from initial to the final position so it can be positive or negative depending on its direction. The displacement can be zero even if the distance is not zero.<\/p>\n Solution 8S.<\/strong><\/span><\/p>\nYes, displacement can be zero even if the distance is not zero.<\/p>\n For example, when a body is thrown vertically upwards from a point A on the ground, after sometime it comes back to the same point A. Then, the displacement is zero, but the distance travelled by the body is not zero (it is 2h; h is the maximum height attained by the body).<\/p>\n Solution 9S.<\/strong><\/span><\/p>\nThe magnitude of displacement is equal to distance if the motion of the body is one-dimensional.<\/p>\n Solution 10S.<\/strong><\/span><\/p>\nThe velocity of a body is the distance travelled per second by the body in a specified direction. \nIts SI unit is metre\/second (m\/s).<\/p>\n Solution 11S.<\/strong><\/span><\/p>\nThe speed of a body is the rate of change of distance with time. \nIts SI unit is metre\/second (m\/s).<\/p>\n Solution 12S.<\/strong><\/span><\/p>\nSpeed is a scalar quantity, while velocity is a vector quantity. The speed is always positive-it is the magnitude of velocity, but the velocity is given a positive or negative sign depending upon its direction of motion. The average velocity can be zero but the average speed is never zero.<\/p>\n Solution 13S.<\/strong><\/span><\/p>\nVelocity gives the direction of motion of the body.<\/p>\n Solution 14S.<\/strong><\/span><\/p>\nInstantaneous velocity is equal to average velocity if the body is in uniform motion.<\/p>\n Solution 15S.<\/strong><\/span><\/p>\nIf a body travels equal distances in equal intervals of time along a particular direction, then the body is said to be moving with a uniform velocity. However, if a body travels unequal distances in a particular direction in equal intervals of time or it moves equal distances in equal intervals of time but its direction of motion does not remain same, then the velocity of the body is said to be variable (or non-uniform).<\/p>\n Solution 16S.<\/strong><\/span><\/p>\nAverage speed is the ratio of the total distance travelled by the body to the total time of journey, it is never zero. If the velocity of a body moving in a particular direction changes with time, then the ratio of displacement to the time taken in entire journey is called its average velocity. Average velocity of a body can be zero even if its average speed is not zero.<\/p>\n Solution 17S.<\/strong><\/span><\/p>\nThe motion of a body in a circular path with uniform speed has a variable velocity because in the circular path, the direction of motion of the body continuously changes with time. \n<\/p>\n Solution 18S.<\/strong><\/span><\/p>\nIf a body starts its motion from a point and comes back to the same point after a certain time, then the displacement is zero, average velocity is also zero, but the total distance travelled is not zero, and therefore, the average speed in not zero.<\/p>\n Solution 19S.<\/strong><\/span><\/p>\nAcceleration is the rate of change of velocity with time. \nIts SI unit is metre\/second2<\/sup>\u00a0(m\/s2<\/sup>).<\/p>\nSolution 20S.<\/strong><\/span><\/p>\nAcceleration is the increase in velocity per second, while retardation is the decrease in velocity per second. Thus, retardation is negative acceleration. In general, acceleration is taken positive, while retardation is taken negative.<\/p>\n Solution 21S.<\/strong><\/span><\/p>\nThe acceleration is said to be uniform when equal changes in velocity take place in equal intervals of time, but if the change in velocity is not the same in the same intervals of time, the acceleration is said to be variable.<\/p>\n Solution 22S.<\/strong><\/span><\/p>\nRetardation is the decrease in velocity per second. \nIts SI unit is metre\/second2<\/sup>\u00a0(m\/s2<\/sup>).<\/p>\nSolution 23S.<\/strong><\/span><\/p>\nVelocity determines the direction of motion.<\/p>\n Solution 24S.<\/strong><\/span><\/p>\n(a) Example of uniform velocity: A body, once started, moves on a frictionless surface with uniform velocity. \n(b) Example of variable velocity: A ball dropped from some height is an example of variable velocity. \n(c) Example of variable acceleration: The motion of a vehicle on a crowded road is with variable acceleration. \n(d) Example of uniform retardation: If a car moving with a velocity ‘v’ is brought to rest by applying brakes, then such a motion is an example of uniform retardation.<\/p>\n Solution 25S.<\/strong><\/span><\/p>\nInitially as the drops are equidistant, we can say that the car is moving with a constant speed but later as the distance between the drops starts decreasing, we can say that the car slows down.<\/p>\n Solution 26S.<\/strong><\/span><\/p>\nWhen a body falls freely under gravity, the acceleration produced in the body due to the Earth’s gravitational acceleration is called the acceleration due to gravity (g). The average value of g is 9.8 m\/s2<\/sup>.<\/p>\nSolution 27S.<\/strong><\/span><\/p>\nNo. The value of ‘g’ varies from place to place. It is maximum at poles and minimum at the Equator on the surface of the Earth.<\/p>\n Solution 28S.<\/strong><\/span><\/p>\nIn vacuum, both will reach the ground simultaneously because acceleration due to gravity is same (=g) on both objects.<\/p>\n Solution 1M.<\/strong><\/span><\/p>\nVelocity is a vector quantity. The others are all scalar quantities.<\/p>\n Solution 2M.<\/strong><\/span><\/p>\nm s-1<\/sup><\/p>\nSolution 3M.<\/strong><\/span><\/p>\nm s-2<\/sup><\/p>\nSolution 4M.<\/strong><\/span><\/p>\nThe displacement is zero.<\/p>\n Solution 5M.<\/strong><\/span><\/p>\n5 m s-1<\/sup><\/p>\nSolution 1N.<\/strong><\/span><\/p>\n<\/p>\n Solution 2N.<\/strong><\/span><\/p>\n<\/p>\n Solution 3N.<\/strong><\/span><\/p>\n<\/p>\n Solution 4N.<\/strong><\/span><\/p>\n18 km h-1<\/sup>\u00a0<\u00a010 m s-1<\/sup>\u00a0<\u00a01 km min-1<\/sup><\/p>\nSolution 5N.<\/strong><\/span><\/p>\nTotal time taken = 3 hours \nSpeed of the train = 65 km\/hr \nDistance travelled = speed x time \n= 65 x 3 = 195 km<\/p>\n Solution 6N.<\/strong><\/span><\/p>\n<\/p>\n Solution 7N.<\/strong><\/span><\/p>\n \n(ii) Average velocity of the train is zero because the train stops at the same point from where it starts, i.e. the displacement is zero.<\/p>\n
Solution 8N.<\/strong><\/span><\/p>\n<\/p>\n Solution 9N.<\/strong><\/span><\/p>\nHere, final velocity = 10 m\/s \nInitial velocity = 0 m\/s \nTime taken = 2s \nAcceleration = (Final Velocity – Initial Velocity)\/time \n= (10\/2) m\/s2 \n<\/sup>\u00a0= 5 m\/s2<\/sup><\/p>\nSolution 10N.<\/strong><\/span><\/p>\nHere, final velocity = 180 m\/s \nInitial velocity = 0 m\/s \nTime taken = 0.05 h or 180 s \nAcceleration = (Final Velocity – Initial Velocity)\/time \n= (180-0)\/180 m\/s2 \n<\/sup>\u00a0= 1 m\/s2<\/sup><\/p>\nSolution 11N.<\/strong><\/span><\/p>\nHere, final velocity = 20 m\/s \nInitial velocity = 50 m\/s \nTime taken = 3 s \nAcceleration = (Final Velocity – Initial Velocity)\/time \n= (20 – 50)\/3 m\/s2 \n<\/sup>\u00a0= -10 m\/s2 \n<\/sup>Negative sign here indicates that the velocity decreases with time, so retardation is 10 m\/s2<\/sup>.<\/p>\nSolution 12N.<\/strong><\/span><\/p>\nHere, final velocity = 18 km\/h or 5 m\/s \nInitial velocity = 0 km\/h \nTime taken = 2 s \nAcceleration = (Final Velocity – Initial Velocity)\/time \n= (5 – 0) \/ 2 m\/s2 \n<\/sup>\u00a0= 2.5 m\/s2<\/sup><\/p>\nSolution 13N.<\/strong><\/span><\/p>\nAcceleration = Increase in velocity\/time taken \nTherefore, increase in velocity = Acceleration \u00d7 time taken \n= (5 \u00d7 2) m\/s \n= 10 m\/s<\/p>\n Solution 14N.<\/strong><\/span><\/p>\nInitial velocity of the car, u = 20 m\/s \nRetardation = 2 m\/s2 \n<\/sup>Given time, t = 5 s \nLet ‘v’ be the final velocity. \nWe know that, Acceleration = Rate of change of velocity \/time \n= (Final velocity – Initial velocity)\/time \nOr, -2 = (v – 20) \/ 5 \nOr, -10 = v – 20 \nOr, v = -20 + 10 m\/s \nOr, v = -10 m\/s \nNegative sign indicates that the velocity is decreasing.<\/p>\nSolution 15N.<\/strong><\/span><\/p>\nInitial velocity of the bicycle, u = 5 m\/s \nAcceleration = 2 m\/s2 \n<\/sup>Given time, t = 5 s \nLet ‘v’ be the final velocity. \nWe know that, acceleration = Rate of change of velocity\/time \n= (Final velocity – Initial velocity)\/time \nOr 2 = (v – 5)\/5 \nOr, 10 = (v – 5) \nOr, v = -5 – 10 \nOr, v = -15 m\/s \nNegative sign indicates that the velocity is decreasing.<\/p>\nSolution 16N.<\/strong><\/span><\/p>\nInitial velocity of the bicycle, u = 18 km\/hr \nTime taken, t = 5 s \nFinal velocity, v = 0 m\/s (As the car comes to rest) \n \n(iii) Let ‘V’ be the speed of the car after 2 s of applying the brakes. \nThen, Acceleration = (V – 5)\/ 2 \nOr, -1 = (V – 5)\/2 \nOr, V = -2 + 5 \nOr, V = 3 m\/s2<\/sup><\/p>\nExercise 2(B)<\/strong><\/span><\/p>\nSolution 1S.<\/strong><\/span><\/p>\nFor the motion with uniform velocity, distance is directly proportional to time.<\/p>\n Solution 2S.<\/strong><\/span><\/p>\nFrom displacement-time graph, the nature of motion (or state of rest) can be understood. The slope of this graph gives the value of velocity of the body at any instant of time, using which the velocity-time graph can also be drawn.<\/p>\n Solution 3S.<\/strong><\/span><\/p>\n(a) Slope of a displacement-time graph represents velocity. \n(b) The displacement-time graph can never be parallel to the displacement axis because such a line would mean that the distance covered by the body in a certain direction increases without any increase in time, which is not possible.<\/p>\n Solution 4S.<\/strong><\/span><\/p>\n(a) There is no motion, the body is at rest. \n(b) It depicts that the body is moving away from the starting point with uniform velocity. \n(c) It depicts that the body is moving towards the starting point with uniform velocity. \n(d) It depicts that the body is moving with variable velocity.<\/p>\n Solution 5S.<\/strong><\/span><\/p>\n<\/p>\n Solution 6S.<\/strong><\/span><\/p>\n(i) The slope of the velocity-time graph gives the value of acceleration. \n(ii) The total distance travelled by a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (without any sign). \n(iii) The displacement of a body in a given time is given by the area enclosed between the velocity-time graph and X-axis (with proper signs).<\/p>\n Solution 7S.<\/strong><\/span><\/p>\nVehicle A is moving with a faster speed because the slope of line A is more than the slope of line B.<\/p>\n Solution 8S.<\/strong><\/span><\/p>\n(a) Fig. 4.33 (a) represents uniformly accelerated motion. For example, the motion of a freely falling object. \n(b) Fig. 4.33 (b) represents motion with variable retardation. For example, a car approaching its destination.<\/p>\n Solution 9S.<\/strong><\/span><\/p>\n \nIn this graph, initial velocity = u \nVelocity at time t = v \nLet acceleration be ‘a’ \nTime = t \nThen, distance travelled by the body in t s = area between the v-t graph and X-axis \nOr distance travelled by the body in t s = area of the trapezium OABD \n= (1\/2) \u00d7 (sum of parallel sides) \u00d7 (perpendicular distance between them) \n= (1\/2) \u00d7 (u + v) \u00d7 (t) \n= (u + v)t \/2<\/p>\n
Solution 10S.<\/strong><\/span><\/p>\nThe slope of the velocity-time graph represents acceleration.<\/p>\n Solution 11S.<\/strong><\/span><\/p>\nCar B has greater acceleration because the slope of line B is more than the slope of line A.<\/p>\n Solution 12S.<\/strong><\/span><\/p>\nFor body A: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration for body A is zero. \nFor body B: The graph is a straight line. So, the slope gives constant velocity. Hence, the acceleration for body B is also zero. \nFor body C: The slope of the graph is decreasing with time. Hence, the acceleration is negative. \nFor body D: The slope of the graph is increasing with time. Hence, the acceleration is positive.<\/p>\n Solution 13S.<\/strong><\/span><\/p>\n \nVelocity-time for a body moving with uniform velocity and uniform acceleration.<\/p>\n
Solution 14S.<\/strong><\/span><\/p>\nRetardation is calculated by finding the negative slope.<\/p>\n Solution 15S.<\/strong><\/span><\/p>\n \nThe area enclosed between the straight line and time axis for each interval of time gives the value of change in speed in that interval of time.<\/p>\n
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