(I) The velocity time graph for an object is shown in the following figure: (3 m)
(a) State the kind of motion that the above graph represents.
(b) What does the slope of the graph represent?
(c) Which is the physical quantity which is measured by the area occupied below the graph?
(d) Calculate the distance travelled by the object in 20 seconds.
(II) Derive the second equation of motion by graphical method. (2 m)
(III) A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds. calculate its acceleration?
Answers
Answer:
The velocity time graph for an object is shown in the following figure: (3 m)
(a) State the kind of motion that the above graph represents.
(b) What does the slope of the graph represent?
(c) Which is the physical quantity which is measured by the area occupied below the graph?
(d) Calculate the distance travelled by the object in 20 seconds.
(II) Derive the second equation of motion by graphical method. (2 m)
(III) A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds. calculate its acceleration?
Explanation:
The velocity time graph for an object is shown in the following figure: (3 m)
(a) State the kind of motion that the above graph represents.
(b) What does the slope of the graph represent?
(c) Which is the physical quantity which is measured by the area occupied below the graph?
(d) Calculate the distance travelled by the object in 20 seconds.
(II) Derive the second equation of motion by graphical method. (2 m)
(III) A car accelerates uniformly from 18 km/h to 36 km/h in 5 seconds. calculate its acceleration?
Answer:
(1) Uniformly accelerated motion, since the slope of the velocity-time graph is positive and uniform.
(2) It represents acceleration.
(3) The area below the velocity-time graph and time axis represents the distance covered.
(4) The area covered in 15s i.e., area of triangle OBC gives the distance covered in 15s.
∴ Distance covered
=
2
1
×BC×OC=
2
1
×30×15Distance=225m.