An object moves with constant acceleration 4.95 m/s2 and over a time interval reaches a final velocity of 13.2 m/s.
(a) If its initial velocity is 6.6 m/s, what is its displacement during the time interval?
(b) What is the distance it travels during this interval?
a) 13.2
13.2^2=6.6^2+2(4.95)(ΔX)
174.24= 43.56+9.9ΔX
130.68=9.9ΔX
ΔX=13.2
b)6.6+13.2/2=13.2/t
9.9=13.2/t
t= 1.33
To solve this problem, we can use the equations of motion under constant acceleration. Recall the following equation:
v = u + at
Where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration, and
- t is the time.
In this case, we are given:
- u = 6.6 m/s,
- a = 4.95 m/s², and
- v = 13.2 m/s.
(a) To find the displacement, we need to use the following equation:
s = ut + (1/2)at²
Substituting the given values, we have:
s = (6.6 m/s)(t) + (1/2)(4.95 m/s²)(t²)
First, we need to find the value of t. Rearranging the equation v = u + at, we have:
t = (v - u)/a
Plugging in the values, we get:
t = (13.2 m/s - 6.6 m/s)/(4.95 m/s²)
Simplifying, we get:
t = 1.3232 s
Now, substituting this value of t back into the equation for displacement, we have:
s = (6.6 m/s)(1.3232 s) + (1/2)(4.95 m/s²)(1.3232 s)²
Solving for s, we get:
s ≈ 4.43 m
Therefore, the object's displacement during the time interval is approximately 4.43 meters.
(b) To find the distance traveled, we need to consider whether the object changes direction during the given time interval. Since the object is moving with a constant acceleration, its velocity is increasing over time, but we are not given any information about changes in direction. Therefore, we can assume that the object moves in a straight line.
In this case, distance traveled can be found by taking the absolute value of the displacement. So:
distance = |s| = |4.43 m| = 4.43 m
Therefore, the object travels a distance of approximately 4.43 meters during this interval.