A herd of wildebeests led by Roger is charging down the hall on the way to lunch. They start out at 10 m/s and accelerate at a constant rate for 5.35 s to reach 34 m/s by the time they reach the stairs. How quickly did Roger's velocity change?
acceleration
=(v1-v0)/t
=(34-10)/5.35 m/s²
=4.486 m/s²
would that be the following equation:
d = 1/2 (Vi + Vf)t
or
Vf^2=Vi^2+ 2ad
nope, Lynn, see MathMate's response.
To find out how quickly Roger's velocity changed, we need to calculate the acceleration.
Acceleration (a) can be calculated using the formula:
a = (final velocity - initial velocity) / time
Given:
Initial velocity (u) = 10 m/s
Final velocity (v) = 34 m/s
Time (t) = 5.35 s
Plugging in the values into the formula:
a = (34 m/s - 10 m/s) / 5.35 s
Simplifying the equation, we get:
a = 24 m/s / 5.35 s
Calculating the value:
a = 4.48 m/s²
Now that we have the acceleration, we can calculate how quickly Roger's velocity changed. Since Roger is part of the herd, we can assume his initial velocity is the same as the herd's initial velocity.
Velocity change (Δv) can be calculated using the formula:
Δv = a * t
Given:
Acceleration (a) = 4.48 m/s²
Time (t) = 5.35 s
Plugging in the values into the formula:
Δv = 4.48 m/s² * 5.35 s
Calculating the value:
Δv = 23.968 m/s
Therefore, Roger's velocity changed by approximately 23.968 m/s.