im having problem understanding what motion equation to use.
" A motercycle moving 12 m/s [W] accelerates at 6.0 m/s [W]. How long will it take to experience a displacement of 63 m [w] ? "
To solve this problem, we can use the motion equation that relates acceleration, initial velocity, time, and displacement. This equation is:
s = ut + (1/2)at^2
Where:
s is the displacement (in meters)
u is the initial velocity (in meters per second)
t is the time (in seconds)
a is the acceleration (in meters per second squared)
Given:
Initial velocity (u) = 12 m/s [W] (westward)
Acceleration (a) = 6.0 m/s^2 [W] (westward)
Displacement (s) = 63 m [W] (westward)
We want to find the time (t) it takes to experience this displacement.
Step 1: Rearrange the equation to solve for time (t):
s = ut + (1/2)at^2
Rearranging, we get:
(1/2)at^2 + ut - s = 0
Step 2: Substitute the given values into the equation:
(1/2)(6)t^2 + (12)t - 63 = 0
Step 3: Solve the quadratic equation using either the quadratic formula or factoring. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
For our equation, a = (1/2)(6) = 3, b = 12, and c = -63. Plugging these values into the quadratic formula:
t = (-12 ± √(12^2 - 4(3)(-63))) / 2(3)
Now, we can simplify the equation:
t = (-12 ± √(144 + 756)) / 6
t = (-12 ± √(900)) / 6
t = (-12 ± 30) / 6
Now, we have two possible solutions:
t1 = (-12 + 30) / 6 = 18 / 6 = 3 seconds
t2 = (-12 - 30) / 6 = -42 / 6 = -7 seconds
Since time cannot be negative in this context, we ignore t2 and conclude that it will take 3 seconds for the motorcycle to experience a displacement of 63 m [W].