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IX CBSE SCIENCE UNIT TEST - SOLUTIONS Motion Test Code: 9SUT101
Marks: 30 Duration: 1 hr
1. What is meant by average velocity of a body moving in a particular direction? (1 Mark) (Ans) If the velocity of a body in a particular direction changes continuously at a uniform rate then the arithmetic mean of the initial and final velocities over a given period of time is called average velocity in that direction. Average velocity in a particular direction, Vav = (Initial velocity + final velocity ) / 2 2. Define the term displacement. (1 Mark) (Ans) The change in the position of a moving body in a particular direction is called its displacement. It is the shortest distance measured from the initial position to the final position of a body. 3. What do the following measure in a car? (a) Speedometer (b) Odometer. (Ans) Speedometer: It measures speed of a running car at any instant of time. Odometer: It measures the distance travelled by a car. 4. A body thrown vertically upward rises up to a height 'h', and comes back to the initial position. Calculate i) the total distance travelled by the body ii) the displacement of the body. (2 Marks) (Ans) The body thrown vertically up rises to a height h and then returns to its initial position.
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i)The total distance travelled = h+h= 2h ii) The body returns to its initial state. So the initial and final position coincides with each other. Therefore, displacement of the body = 0(zero). 5. A train starting from rest attains a velocity of 72 km h-1 in 5 min. Assuming that the acceleration is uniform, find (i) The acceleration (ii) The distance travelled by the train for attaining this velocity. (2 Marks) (Ans) Given,Initial velocity,u = 0 (rest); v = 72 km/h = 20 ms-1; t = 5 min = 300 s (i) Fromfirst equation of motion, v = u + at or, a = (v-u)/t or, = [20 ms-1- 0 ms-1] / 300 or, a = 1/15 ms-2 (ii) Using third equation of motion, v2 –u2 = 2as or, s = (v2 –u2)/2a a. = [(20 ms-1) 2- (0 ms-1) 2] / 2 x (1/15) ms-2 = 3000 m = 3 km The acceleration of the train is 1/15 ms-2 & the distance traveled is 3 km. 6. A moving train is brought to rest within 20 sec by applying brakes. Find the initial velocity, if the retardation due to brakes is 2 m/s2. (2 Marks) (Ans) Final velocity, v = 0 (rest) Since, Retardation = -ve acceleration Therefore, a = - 2 m/s2 Using the first equation of motion, v = u + at Subsituting the given values, 0 = u + (-2) × 20 or, 0 = u – 40 or, u = 40 ms-1 7. Acceleration is the rate of change of velocity. Draw a graph to show the motion of a body with uniform acceleration. How can you find the acceleration from that graph? (2 marks)
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(Ans) Graph for the uniform acceleration is given below:
Acceleration can be found by finding the slope of the line. 8. The graph shows the position of body at different times. Calculate the speed of the body as it moves
from: (i) (ii) (iii)
A to B B to C and C to D.
(3 Marks)
(Ans) (i) For A to B Speed= (Distance on Y axis / Time on X axis) b. = (3.0 m / 6.0 s) c. = 0.5 m/s (ii) There is no movement during BC part because slope is parallel to time axis. Hence the speed in this part will be zero. (iii) For CD Distance = (7 – 3)m = 4 m Time = (12 – 10)s = 2 s Speed= (Distance on Y axis / Time on X axis) d. = (4.0 m / 2.0 s)
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e. = 2.0 m/s 9. A man travels a distance of 1.5m towards East, then 2.0m towards South and finally 4.5m towards East. i) What is the total distance traveled? ii) What is his total displacement? (3 Marks) (Ans) i) Total distance travelled = 1.5+2.0+4.5=8.0m ii) His resultant displacement can be explained by drawing a map
Let 1 cm represent 1m Then man travels 1.5 cm from A to B toward east, then 2.0cm from B to C towards south and finally C to D 4.5 cm towards east.Resultant displacement can be found by joining the starting point A with the finishing Point D. Thenthe line AD represents the final resultant displacement of the man. On measuring, we find AD = 6.3 cm Since 1 cm = 1metre 6.3 cm = 6.3metre Thus, the resultant displacement of the man is 6.3 metres 10. A racing car has a uniform acceleration of 4m/s2. What distance will it cover in 10 seconds after the start? (3 Marks) (Ans) Given, initial velocity, u = 0 (i) Time t = 10 s (ii) Acceleration, a = 4 m/s2 And Distance s = ? By second equation of motion: s = ut + (1/2) at2 s = 0 + (1/2) x 4 x (10)2 s = 200 m Thus, the distance covered by the car in 10 seconds is 200 m. 11. Derive the three equations of motion from the speed-time graph. (5 Marks) (Ans) The speed time (or velocity-time) graph for a body under uniform acceleration is given in the fig. as:
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Derivation of first equation of motion: Equation for velocity-time relation: According to definition, Acceleration of a body = Rate of change of velocity = slope of velocity-time graph The velocity-time graph for a uniformly accelerated body is given by the straight line AB. So, acceleration of the body is equal to the slope of the line AB. Thus, Acceleration = Slope of line AB = BD/AD = (BC-DC)/AD From the graph, BC= v, DC= OA =u, AD=OC=t Then, a = (v-u)/t or, at = v-u or, v = u + at Derivation of second equation of motion: Equation for position-time relation – Distance travelled, s = area of the trapezium OABC or, s=1/2 (area of the triangle ABD) + area of the rectangle OADC or, s = (1/2 × base × height) + ( length × breadth) s= (1/2 ×AD ×BD) + (OC × OA)……………………(1) From the figure, AD = OC=t BD = (BC-DC) = (v-u) = (u + at-u) = at OA = u Putting these above values in equation (1), we get s = (1/2 × t × at) + (t × u ) s = u t + 1/2 a t2
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This is the second equation of motion. Derivation of the third equation of motion: equation for position–velocity relation Distance travelled, s = area of the trapezium OABC s = ½ (sum of the parallel sides) × perpendicular distance between the two parallel sides or, s = ½ (OA + BC) × OC Putting the values of OA, BC and OC, we get s = ½ (u + v) × t …………………………………(2) As we know, v = u + at or, t = (v-u)/a Putting the value of t in equation (2), we get s = ½ (u + v)× ((v-u)/a) or, 2as = (u + v)(v-u) = v2 – u2 or, 2as = v2 – u2 This is the third equation of motion. 12. Two trains X and Y are running on parallel tracks with a speed of 72km/h and 54km/h respectively. The driver of train X applies the brake and it comes to a stop in 10seconds. While the driver of train Y applies the brake and the train retards uniformly before coming to rest in 15 seconds. Plot distance-time graphs for both the trains. Also, calculate the distance travelled by each train after the brakes were applied. (5 Marks) (Ans) For train X: u = 72km/h = [72 x (5/18)]m/s = 20m/s v = 0; t = 10s For train Y: u = 54km/h = [54 x (5/18)]m/s = 15m/s v = 0; t = 15s Speed-time graph for trains X and Y is:
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