2. A partial record of a free falling body showed only the three consecutive points. The distance from the first to the second was 24.25 cm. that from the first to the third was 58.92 cm. The time interval was 1/10. Find the acceleration and the instantaneous speed of the body at the time of the recorded point.
To find the acceleration and instantaneous speed of the body, we can use the equations of motion. Here's how you can calculate it step by step:
Step 1: Calculate the time interval between the first and second points
The time interval is given as 1/10, which represents 1/10th of a second. So, the time interval (t) between the first and second points is:
t = 1/10
Step 2: Calculate the distance covered between the first and second points
The distance between the first and second points is given as 24.25 cm.
Step 3: Calculate the acceleration
The equation to calculate the acceleration (a) is:
a = (2 * (d - v0 * t)) / (t^2)
Where:
d is the distance covered between the first and second points (24.25 cm)
v0 is the initial velocity (0 for a free-falling body)
t is the time interval (1/10)
Since the initial velocity of a free falling body is 0, the equation simplifies to:
a = (2 * d) / (t^2)
Substituting the given values:
a = (2 * 24.25) / ((1/10)^2)
a = 485 / (1/100)
a = 48500 cm/s^2
Therefore, the acceleration of the body is 48500 cm/s^2.
Step 4: Calculate the instantaneous speed
The equation to calculate the instantaneous speed (v) is:
v = v0 + a * t
Substituting the given values:
v = 0 + 48500 * (1/10)
v = 4850 cm/s
Therefore, the instantaneous speed of the body at the time of the recorded point is 4850 cm/s.
So, the acceleration of the body is 48500 cm/s^2 and the instantaneous speed at the recorded point is 4850 cm/s.