question_answer

  Derive the equations of motion for uniformly accelerated motion from velocity-time graph. Or Distance travelled by a train and time taken by it is shown in the following table. (a) Plot distance-time graph. (b) What is the average speed of the train? (c) When is the train travelling at the highest speed? (d) At what distance does the train slowdown? (e) Calculate the speed of the train between 10:40 AM to 11:00 AM.

Time	Distance (in km)
10:00 am	0
10:30 am	25
10:40 am	28
11:00 am	40
11:15 am	42
11:30 am	50

 

Answer:

  Equations of motion by graphical method Consider an object moving along a straight line with an initial velocity u and uniform acceleration a. Suppose, it travels adistance, s in time, t. As shown in figure its velocity-time graph is a straight line. Here,                                                                                                                 (1)                                                                                                                                                              (1) (i) Equation for Velocity-time Relation We know that Acceleration = Change in Velocity/Time This proves the first equation of motion.                                                                                                                              (1) (ii) Equation for Position-time Relation Distance travelled by an object in time t is s. s = Area of the trapezium OABE = Area of OADE + Area of ADB or...(i) Now, DB = BE - DE = v - u = at Putting this value for DB in Eq. (i), weget This proves the second equation of motion.                                                                                                                        (1) (iii) Equation for Position-velocity Relation The distance travelled by an object in time t is s = Area of the trapezium OABE Substituting EB, ED and OE with v, u and t, we get   ...(ii) But from the first equation of motion, we know that or                            Substituting t in Equation (ii) with this value, we get or                                                                                                                                                                                                 (1) This proves the third equation of motion.                                                                                                                             (1) Or (i) Distance-time graph          (ii) Average speed In this problem, total distance travelled = 50 km Total time taken 10:00 AM to 11:30 AM = 1 hour 30 minutes Now average speed                 (iii) We know, speed = slope of distance-time graph. The greater the slope, the greater is the speed. From the graph it is clear that slope of distance-time graph is maximum between 10:00 AM to 10:30 AM, so thetrain was travelling at the highest speed during this interval of time. (iv) The part CD of the graph has minimum slope, so the train had minimum speed between 11:00 AM and 11:15 AM, Thus, the train had slowed down between 40 km and 42 km. (v) Speed between 10:40 AM to 11:00 AM                                                                                                  (1x5)