A motorboat starting from rest on a lake accelerates

  1. A man is 48 m behind a bus which is at rest. The bus starts accelerating at the rate of 1 m/s^2 , at the same time the man starts running with uniform velocity of 10 m/s . What is the minimum time in which the man catches the bus?
  2. Second Equation of Motion
  3. [Solved] A motorboat starting from rest on a lake accelerates in a st
  4. CBSE Class IX ( 9th) Science


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A man is 48 m behind a bus which is at rest. The bus starts accelerating at the rate of 1 m/s^2 , at the same time the man starts running with uniform velocity of 10 m/s . What is the minimum time in which the man catches the bus?

Given, Acceleration of bus, a = 1 m s − 2 Assume man catches the train in time t. Displacement by man in time t s 1 ​ = v t = 1 0 t Displacement by bus in time t, s 2 ​ = u t + 2 1 ​ a t 2 = 2 t 2 ​ Separation between man and bus is 4 8 m s 1 ​ = s 2 ​ + 4 8 1 0 t = 2 t 2 ​ + 4 8 t 2 − 2 0 t + 9 6 = 0 t = 2 2 0 ± 2 0 2 − 4 × 9 6 ​ ​ t = 8 a n d 1 2 In first attempt he caches the bus Hence, minimum time he catches the bus is 8 sec

Second Equation of Motion

Questions Q 1 Page 109 - A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled Q 4 Page 110 - A racing car has a uniform acceleration of 4 m s-2. What distance will it cover in 10 s after start? NCERT Question 4 - A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s–2 for 8.0 s. How far does the boat travel during this time? Example 8.6 - A car accelerates uniformly from 18 km h–1 to 36 km h – 1 in 5 s.Calculate (i) the acceleration and (ii) the distance covered by the car in that time. Example 8.7 - The brakes applied to a car produce an acceleration of 6 m s-2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time. Transcript Second Equation of Motion - Derivation s = ut + 1/2 𝑎𝑡^2 Derivation 𝜋× We know that Velocity = 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡/𝑇𝑖𝑚𝑒 Velocity × Time = Displacement Displacement = Velocity × Time If Velocity is not constant (i.e. Velocity keeps on increasing or decreasing) We can also take Average Velocity in place of Velocity So our formula becomes Displacement = Average Velocity × Time Displacement = (𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 + 𝐹𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦)/2 × Time s = ((𝑢 + 𝑣)/2) × t From first equation of motion, we know that: v = u + at Putting value of v in this equation s = ((𝑢 + (𝒖 + 𝒂𝒕))/2) × t s = ((2𝑢 + 𝑎𝑡)/2) × t s = (2𝑢/2+𝑎𝑡/2) × t s = (𝑢+1/2 𝑎𝑡...

[Solved] A motorboat starting from rest on a lake accelerates in a st

The correct answer is 96 m • Given, the initial velocity of the boat = 0 m/s • Acceleration of the boat = 3 m/s 2 • Time period = 8s • As per the second motion equation, s = ut + 1/2 at 2 • Therefore, the total distance traveled by boat in 8 seconds = 0 + 1/2 [(3) ×(8) 2] • s = 96 meters • Therefore, the motorboat travels a distance of 96 meters in a time frame of 8 seconds. ​​ Important Points • There are three equations of motion that can be used to derive components such as displacement(s), velocity (initial and final), time(t) and acceleration(a). • The following are the three equation of motion: • First Equation of Motion: v = u + at • Second Equation of Motion: s = ut + 1/2(at 2) • Third Equation of Motion: v 2= u 2+ 2as

CBSE Class IX ( 9th) Science

Page 100 ( CBSE Class IX ( 9th) Science Textbook - Chapter 8. Motion ) Question 1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example. Answer : Yes, an object even after it has moved through a distance, it can have zero displacement. As we know distance is just length of the path an object has covered irrespective of its direction or position with reference to certain point, where as the shortest distance measured from the initial to the final position of an object is known as the displacement. For example, an object starts from point A and after covering a distance of say 50 meters, reaches at point B. Here after, it again moves back to point A. Here the distance covered by object is = AB + BA = 50 m + 50 m = 100 m where as displacement of object is = AB - BA = 50 m - 50 m = 0 m As initial position of object is same as that of its final position hence its displacement, which is distance measured from the initial to the final position, is zero. A >-----------50 m-------------> <-----------50 m-------------< B 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 from his initial position? Answer : Suppose, a farmer moves along the boundary of a square field of side 10 m in 40 s as shown in the figure given below. Distance cover by the farmer as he moves from A to B to C to D to A, along t...

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