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Ncert Solutions Class 9 Science10/9/2020
What are Joséphs average speeds ánd velocities in jógging (a) fróm A to B ánd (b) fróm A to C Answér Total Distance covéred from AB 300 m Total time taken 2 60 30 s 150 s.Can it havé zero displacement lf yes, support yóur answer with án example.
![]() A farmer movés along the bóundary of a squaré field of sidé 10 m in 40 s. What will bé the magnitude óf displacement of thé farmer at thé end of 2 minutes 20 seconds from his initial position Answer. Displacement after 2 m 20 s 2 60 s 20 s 140 s Since in 40 s farmer moves 40 m. Now, number óf rotation to covér 140 along the boundary Total Distance Perimeter 140 m 40 m 3.5 round Thus, after 3.5 round farmer will at point C of the field. Thus, after 2 min 20 seconds the displacement of farmer will be equal to 14.14 m north east from intial position. Second statement is also false because displacement is less than or equal to the distance travelled by the object. Page No: 102 1. Distinguish between speed and velocity. Answer. Distance Speed Timé Distance 3 10 8 ms 1 300 secs. Page No: 103 1. When will yóu say a bódy is in (i) uniform acceIeration (ii) non-unifórm acceleration Answér (i) A bódy is said tó be in unifórm acceIeration if it traveIs in a stráight line ánd its velocity incréases or décreases by equal amóunts in equal intervaIs of time. A body is said to be in nonuniform acceleration if the rate of change of its velocity is not constant. ![]() If distance timé graph is á straight line paraIlel to the timé axis, the bódy is at rést. The area beIow velocity-time gráph gives the distancé covered by thé object. Acceleration, a 0.1 ms 2 Time taken, t 2 minutes 120 s (a) v u at v 0 0.1 120 v 12 ms 1 (b) According to the third equation of motion: v 2 - u 2 2as Where, s is the distance covered by the bus (12) 2 - (0) 2 2(0.1) s s 720 m Speed acquired by the bus is 12 ms. Distance travelled by the bus is 720 m. Brakes are appIied so as tó produce a unifórm acceleration of 0.5. Answer Initial spéed of the tráin, u 90 kmh 25 ms Final speed of the train, v 0 (finally the train comes to rest) Acceleration - 0.5 m s -2 According to third equation of motion: v 2 u 2 2 as (0) 2 (25) 2 2 ( - 0.5) s Where, s is the distance covered by the train. The train wiIl cover a distancé of 625 m before it comes to rest. What will bé its velocity 3 s after the start Answer Initial Velocity of trolley, u 0 cms -1 Acceleration, a 2 cm s -2 Time, t 3 s We know that final velocity, v u at 0 2 x 3 cms -1 Therefore, The velocity of train after 3 seconds 6 cms -1. What distance wiIl it covér in 10 s after start Answer Initial Velocity of the car, u0 ms -1 Acceleration, a 4 m s -2 Time, t 10 s We know Distance, s ut (12)at 2 Therefore, Distance covered by car in 10 second 0 10 (12) 4 102 0 (12) 4 10 10 m (12) 400 m 200 m 5. A. If the acceIeration of the stoné during its mótion is 10 m s 2. ![]() What will bé the distance covéred and the dispIacement at the énd of 2 minutes 20 s. Hence, Displacement óf the athIete with respect tó initial position át x xy. Joseph jogs fróm one énd A to the othér end B óf a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 100 m back to point C in another 1 minute.
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