Advertisements
Advertisements
प्रश्न
A body at rest is thrown downward from the top of the tower. Draw a distance – time graph of its free fall under gravity during the first 3 seconds. Show your table of values starting t = 0 with an interval of 1 second, (g = 10 ms−2).
Advertisements
उत्तर
Initial velocity = M = 0
Acceleration = a = +g = 10 ms−2
when t = Is, then distance travelled (S), is given by
S1 = ut + `1/2` at2
S1 = `0(1)+1/2(10)(1)^2`
S1 = 5m
When t = 2s then S2 = ut + `1/2` at2
S2 = `(0)(2)+1/2(10)(2)^2`
S2 = 5(4) = 20 m
When t = 3s, then S3 = ut + `1/2` at2
S3 = `(0)(3)+1/2(10)(3)^2`
S = 5 (9) = 45 m
| Time | 1s | 2s | 3s |
| Distance covered | 5 m | 20 m | 45 m |
APPEARS IN
संबंधित प्रश्न
What is the quantity which is measured by the area occupied below the velocity-time graph?
Draw a displacement-time graph for a boy going to school with uniform velocity.
Diagram shows a velocity – time graph for a car starting from rest. The graph has three sections AB, BC and CD.
Compare the distance travelled in section BC with the distance travelled in section AB.
A ball is thrown up vertically and returns back to thrower in 6 s. Assuming there is no air friction, plot a graph between velocity and time. From the graph calculate
- deceleration
- acceleration
- total distance covered by ball
- average velocity.
From the diagram given below, calculate distance covered by body.
From the velocity – time graph given below, calculate Distance covered in the region ABCE.
Draw the following graph:
Distance versus time for a body at rest.
Draw velocity-time graph to show:
Deceleration
Write a sentence to explain the shape of graph.
State whether true or false. If false, correct the statement.
The velocity – time graph of a particle falling freely under gravity would be a straight line parallel to the x axis.
The velocity-displacement graph describing the motion of a bicycle is shown in the figure.
The acceleration-displacement graph of the bicycle's motion is best described by:
