Physics
A ball moves on a smooth floor in a straight line with a uniform velocity 10 m s-1 for 6 s. At t = 6 s, the ball hits a wall and comes back along the same line to the starting point with same speed. Draw the velocity-time graph and use it to find the total distance travelled by the ball and its displacement.
Motion in One Dimension
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Answer
The velocity-time graph for the motion of ball on a smooth floor in a straight line is shown below:
![A ball moves on a smooth floor in a straight line with a uniform velocity 10 m s <sup>-1</sup> for 6 s. At t = 6 s, the ball hits a wall and comes back along the same line to the starting point with same speed. Draw the velocity-time graph and use it to find the total distance travelled by the ball and its displacement. Motion in one dimension, Concise Physics Solutions ICSE Class 9.](https://cdn1.knowledgeboat.com/img/cp9/velocity-time-graph-ball-smooth-floor.png)
Total distance travelled = distance travelled in the first 6 s + distance travelled in next 6 s.
As we know,
Distance = velocity x time
Substituting the values in the formula above we get,
Distance when the ball moves towards the wall = 10 m s-1 x 6 s = 60 m
Distance when the ball moves away from the wall = 10 m s-1 x 6 s = 60 m
Total distance = 60 m + 60 m = 120 m
Displacement = distance travelled in the forward direction - distance travelled while coming back
= 60 m - 60 m = 0
ā“ Displacement = 0 as the ball comes back to the initial place.
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Related Questions
Figure below shows a velocity-time graph for a car starting from rest. The graph has three parts AB, BC and CD.
(i) State how is the distance travelled in any part determined from this graph.
(ii) Compare the distance travelled in part BC with the distance travelled in part AB.
(iii) Which part of graph shows motion with uniform (a) velocity (b) acceleration (c) retardation?
(iv) (a) Is the magnitude of acceleration higher or lower than that of retardation? Give a reason. (b) Compare the magnitude of acceleration and retardation.
The velocity-time graph of a moving body is given below in figure
Find ā
(i) The acceleration in parts AB, BC and CD.
(ii) Displacement in each part AB, BC, CD, and
(iii) Total displacement.
Figure shows the velocity-time graph of a particle moving in a straight line.
(i) State the nature of motion of particle.
(ii) Find the displacement of particle at t = 6 s.
(iii) Does the particle change its direction of motion?
(iv) Compare the distance travelled by the particle from 0 to 4 s and from 4 s to 6 s.
(v) Find the acceleration from 0 to 4 s and retardation from 4 s to 6 s.
The correct equation of motion is ā
- v = u + at
- S = ut + at2
- v2 = u2 + 2aS
- All of these.