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
Name the two quantities, the slope of whose graph give acceleration.
A body is moving uniformly in a straight line with a velocity of 5 m/s. Find graphically the distance covered by it in 5 seconds.
What type of motion is represented by the following graph ?

What does the slope of velocity-time graph represent?
Draw velocity – time graph for the following situation:
When a body is moving with variable velocity, but uniform acceleration.
What can you conclude if the speed-time graph of a body is a curve moving upwards starting from the origin?
Given below are the speed -time graphs. Match them with their corresponding motions :
![]() |
(a) Uniformity retared motion |
![]() |
(b) Non-uniformity acceleration |
![]() |
(c) Non-uniform motion |
![]() |
(d) uniform motion |
Which of the following graphs represents a motion with negative acceleration?
Interpret the following graph:
From the v-t graph, ______ can be calculated.




