Consider a cart moving along the x-axis at -2.0 m/s. After a brief acceleration at a rate of -3.0 m/s2 , its velocity is -6.0 m/s. For how long was the cart accelerating in m/s2?
To solve this problem, we can use the equation of motion relating velocity, initial velocity, time, and acceleration:
v = u + at
where:
v = final velocity = -6.0 m/s
u = initial velocity = -2.0 m/s
a = acceleration (unknown)
t = time (unknown)
Since the cart initially moves at -2.0 m/s and accelerates to -6.0 m/s, the change in velocity is (-6.0) - (-2.0) = -4.0 m/s.
Substituting the given values into the equation, we have:
-6.0 m/s = -2.0 m/s + a * t
Rearranging the equation, we get:
-4.0 m/s = a * t
Now, we need to determine the time of acceleration (t) in seconds.
Dividing both sides by a, we have:
t = -4.0 m/s / a
Since we know that the acceleration is given as -3.0 m/s^2, we substitute this value into the equation:
t = -4.0 m/s / (-3.0 m/s^2)
Simplifying:
t = 4/3 s
Therefore, the cart was accelerating for 4/3 seconds, or approximately 1.33 seconds.