To find the distance traveled by a body thrown vertically upward with an initial speed of 20 m/s, under the effect of gravity (g = 10 m/s²) in the first second of its motion, we can use the kinematic equation for distance under constant acceleration (deceleration in this case, because gravity acts downward):
[tex]\[ \text{distance} = v_0 t + \frac{1}{2} a t^2 \][/tex]
Here, [tex]\( v_0 \)[/tex] is the initial speed, [tex]\( t \)[/tex] is the time, and [tex]\( a \)[/tex] is the acceleration. Note that in this problem, gravity is decelerating the object, so [tex]\( a = -g \)[/tex].
Given:
- Initial speed, [tex]\( v_0 = 20 \)[/tex] m/s
- Time, [tex]\( t = 1 \)[/tex] second
- Acceleration due to gravity, [tex]\( a = -10 \)[/tex] m/s²
Now substitute these values into the formula:
[tex]\[ \text{distance} = (20 \text{ m/s}) (1 \text{ second}) + \frac{1}{2} (-10 \text{ m/s}^2) (1 \text{ second})^2 \][/tex]
First, calculate the distance due to the initial speed:
[tex]\[ 20 \text{ m/s} \times 1 \text{ second} = 20 \text{ meters} \][/tex]
Next, calculate the distance due to the acceleration (gravity):
[tex]\[ \frac{1}{2} \times (-10 \text{ m/s}^2) \times (1 \text{ second})^2 = \frac{1}{2} \times (-10) \times 1 = -5 \text{ meters} \][/tex]
The total distance traveled is the sum of these two contributions:
[tex]\[ 20 \text{ meters} + (-5 \text{ meters}) = 15 \text{ meters} \][/tex]
Therefore, the distance traveled by the body in the first second is:
[tex]\[ \boxed{15 \text{ meters}} \][/tex]
So, the correct answer is:
(2) 15 m
