A stone is thrown vertically upwards with a velocity 30 ms −1 . If the acceleration

due to gravity is 10 ms −2 , what is the distance travelled by the particle during the
first second of its motion?

25 m

To find the distance traveled by the particle during the first second of its motion, we need to calculate the total displacement during that time interval.

Step 1: Calculate the maximum height reached by the stone.
To find the maximum height reached by the stone, we can use the following formula:

v^2 = u^2 + 2as

where:
v = final velocity (0 m/s, since the stone reaches the highest point and momentarily stops)
u = initial velocity (30 m/s)
a = acceleration due to gravity (-10 m/s^2)
s = displacement or height reached

Rearranging the equation, we get:
s = (v^2 - u^2) / (2a)

Plugging in the values, we have:
s = (0^2 - 30^2) / (2 * -10)
s = (-900) / (-20)
s = 45 meters

Step 2: Calculate the distance traveled during the first second.
Since the stone is thrown vertically upwards, the distance traveled during the first second will be equal to the distance traveled upwards until it reaches its maximum height.

The time taken to reach the maximum height can be calculated using the equation:
v = u + at

where:
v = final velocity (0 m/s)
u = initial velocity (30 m/s)
a = acceleration due to gravity (-10 m/s^2)
t = time taken

Rearranging the equation, we get:
t = (v - u) / a
t = (0 - 30) / -10
t = 3 seconds

However, we are only interested in the distance traveled during the first second, so we need to find the distance traveled during this time.

Using the equation of motion:
s = ut + (1/2)at^2

where:
s = displacement or distance traveled
u = initial velocity (30 m/s)
t = time (1 second)
a = acceleration due to gravity (-10 m/s^2)

Plugging in the values, we have:
s = 30 * 1 + (1/2) * (-10) * (1)^2
s = 30 - 5
s = 25 meters

Therefore, the stone travels a distance of 25 meters during the first second of its motion.

To find the distance traveled by the stone during the first second of its motion, we need to calculate the height it reaches and then subtract the initial position from it.

To calculate the height reached by the stone, we can use the following kinematic equation:

s = ut + 0.5 * a * t^2

Where:
s = height reached
u = initial velocity = 30 m/s (given)
a = acceleration due to gravity = -10 m/s^2 (negative because it acts in the opposite direction)
t = time = 1 second (first second of motion)

Plugging in the values:

s = (30 * 1) + (0.5 * -10 * 1^2)
s = 30 - 5
s = 25 meters

So, the stone reaches a height of 25 meters during the first second of its motion.

However, we need to calculate the distance traveled, which involves the stone's initial position. Since the stone is thrown vertically upwards, we can assume its initial position is at the ground level (which is usually taken as zero).

Thus, the distance traveled by the stone during the first second of its motion is equal to the height reached, which is 25 meters.