A motorcycle, initially traveling at a speed of 14 m/s, slows down (decelerates) at a rate of -3.5 m/s2 for 2 s. How far does it travel during that time?
d=Vi*t-3.5*1/2*t^2
To find the distance traveled by the motorcycle, we can use the equation of motion:
s = ut + (1/2)at^2
Where:
s = distance traveled
u = initial velocity
t = time taken
a = acceleration (deceleration in this case)
Given:
Initial velocity (u) = 14 m/s
Deceleration (a) = -3.5 m/s^2
Time taken (t) = 2 s
Let's substitute the values into the equation:
s = (14 m/s)(2 s) + (1/2)(-3.5 m/s^2)(2 s)^2
s = 28 m + (1/2)(-3.5 m/s^2)(4 s^2)
s = 28 m - (1/2)(-3.5 m/s^2)(16 s^2)
s = 28 m - (1/2)(-3.5 m/s^2)(16 s^2)
s = 28 m - (-28 m)
s = 28 m + 28 m
s = 56 m
Therefore, the motorcycle travels a distance of 56 meters during that time.
To calculate the distance traveled by the motorcycle, we can use the formula:
distance = initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity (u) = 14 m/s
Deceleration (a) = -3.5 m/s^2 (negative sign indicates deceleration)
Time (t) = 2 s
Plugging the values into the formula, we get:
distance = 14 m/s * 2 s + (1/2) * (-3.5 m/s^2) * (2 s)^2
Simplifying the equation:
distance = 28 m + (1/2) * (-3.5 m/s^2) * 4 s^2
distance = 28 m - 7 m
distance = 21 m
Therefore, the motorcycle travels a distance of 21 meters during the 2-second period.