In an archery practise an arrow was released from the bow with a velocity 20m/s.it strikes to the board situated 15m ahead and stops.find at what rate the arrow was accelerated?
a=(V^2-Vo^2)/2d=(0-(20^2))/30=-13.3m/s^2
Good
To find the rate at which the arrow was accelerated, we need to use the equation of motion that relates velocity, distance, and acceleration.
The equation we can use is: v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s since the arrow stops)
u = initial velocity (20 m/s)
a = acceleration (unknown)
s = distance traveled (15 m)
Rearranging the equation, we get: a = (v^2 - u^2) / (2s)
Plugging in the values we know, we have: a = (0^2 - 20^2) / (2 * 15)
a = (-400) / 30
a = -13.33 m/s^2
Therefore, the rate at which the arrow was accelerated is -13.33 m/s^2. Note that the negative sign indicates that the arrow underwent deceleration or slowing down.