A baseball thrown at 25.0m/s strikes a catcher's mitt and slows down to rest in 0.500 s. What is the magnitude of the ball's acceleration?
I'm having trouble finding the process of figuring out this question. Help?
So you just make the deacceleration the time?
All you need to answer this is the DEFINITION of acceleration.
The acceleration is the speed change divided by the time. In this case, it is a negative number because the speed is decreasing.
a = (0 - 25.0)m/s /0.500 s = ___ m/s^2
Huh? a= -25/.5=-50m/s^2
(CHANGE IN VELOCITY} /TIME
as he said
To find the magnitude of the ball's acceleration, we can use the following equation:
acceleration (a) = change in velocity (Δv) / time taken (Δt)
In this case, the initial velocity is 25.0 m/s, and the final velocity is 0 m/s since the ball comes to rest. The time taken is 0.500 s. Let's substitute these values into the equation:
a = (0 - 25.0) m/s / 0.500 s
Next, we calculate the change in velocity:
Δv = 0 - 25.0 m/s
= -25.0 m/s
Now, we substitute this value into the equation:
a = -25.0 m/s / 0.500 s
Dividing -25.0 m/s by 0.500 s gives us:
a = -50 m/s²
The magnitude of the ball's acceleration is 50 m/s².