A skateboard rider starts from rest and maintains a constant acceleration of +0.50 m/s squared for 8.4 s. What is the rider's displacement during this time?
(1/2) a t^2
4.2m
25 M
To find the rider's displacement during this time, we can use the equation of motion:
s = ut + 0.5at^2
Where:
s = displacement
u = initial velocity (which is 0 m/s since the rider starts from rest)
a = acceleration (+0.50 m/s^2)
t = time interval (8.4 s)
Plugging in the values into the equation, we get:
s = (0)(8.4) + 0.5(0.50)(8.4)^2
Now, we can solve the equation to find the displacement. Let's break it down step by step:
s = (0)(8.4) + 0.5(0.50)(8.4)(8.4)
s = 0 + 0.5(0.50)(70.56)
s = 0 + 0.5(0.25)(70.56)
s = 0 + 0.125(70.56)
s = 0 + 8.82
s = 8.82 meters
Therefore, the skateboard rider's displacement during this time is 8.82 meters.