A speedboat increases in speed from 21 m/s to 30.9 m/s in a distance 191 m.
a)Find the magnitude of its acceleration.
Answer in units of m/s^2.
b)Find the time it takes the boat to travel this distance.
Answer in units of s.
a. a = (V^2-Vo^2)/2d
a = (30.9^2-21^2)/382 = 1.35 m/s^2.
b. 21 + 1.35t = 30.9
1.35t = 9.9
t = 7.33 s.
To find the magnitude of acceleration, we can use the formula:
acceleration (a) = change in velocity (Δv) / time taken (Δt)
Given:
Initial velocity (u) = 21 m/s
Final velocity (v) = 30.9 m/s
Distance (s) = 191 m
a) To find the magnitude of acceleration, we need to determine the change in velocity. Therefore:
Δv = v - u = 30.9 m/s - 21 m/s = 9.9 m/s
Now, let's plug these values into the formula for acceleration:
a = Δv / Δt
To find the time taken (Δt), we can use the following equation:
s = ut + (1/2)at^2
In this case, we know the initial velocity (u), distance (s), and acceleration (a). We're solving for time (t), so we rearrange the equation:
s = ut + (1/2)at^2
191 m = (21 m/s)t + (1/2)a t^2
Since we are solving for time (t), we need to rearrange this equation in terms of t:
(1/2)at^2 + (21 m/s)t - 191 m = 0
This is a quadratic equation. We can solve it using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, the terms a, b, and c correspond to the coefficients of the quadratic equation. Let's substitute them in:
a = (1/2)a
b = (21 m/s)
c = -191 m
Now, we can substitute these values into the quadratic formula to find the time taken (t).
Finally, let's substitute the values into the formula and calculate the time taken.