A ball initially at rest rolls down a hill with
an acceleration of 4.5 m/s2.
If it accelerates for 7.9 s, how far will it
move?
Answer in units of m
Vo = 0
a = 4.5 m/s^2
t = 7.9 s.
d = ?
V = Vo + a*t = 0 + 4.5*7.9 = 35.6 m/s.
V^2 = Vo^2 + 2a*d = 35.6^2
Vo^2 + 2a*d = 35.6^2. Solve for d.
Well, if the ball is rolling down a hill and it's accelerating at 4.5 m/s2, I'd say it's in quite a hurry! I hope it has a valid reason to rush down like that. Anyway, let's calculate how far it will move during its escapade.
To find the distance the ball will move, we can use the equation: distance = (1/2) * acceleration * time^2. Plugging in the values we have:
distance = (1/2) * 4.5 m/s2 * (7.9 s)^2
Now let's do the math:
distance = (1/2) * 4.5 m/s2 * (62.41 s^2)
distance = 140.47 m
So, the ball will move approximately 140.47 meters. That's quite a downhill adventure! Just make sure to stay out of its way and maybe grab some popcorn to enjoy the show.
To find out how far the ball will move, we can use the equation:
distance = initial velocity × time + 0.5 × acceleration × time^2
Given:
Initial velocity (u) = 0 m/s (since the ball is initially at rest)
Acceleration (a) = 4.5 m/s^2
Time (t) = 7.9 s
Plugging these values into the equation, we get:
distance = 0 × 7.9 + 0.5 × 4.5 × (7.9)^2
Simplifying the equation, we have:
distance = 0 + 0.5 × 4.5 × (7.9)^2
distance = 0 + 0.5 × 4.5 × 62.41
distance = 0 + 2.25 × 62.41
distance = 140.4225 m
Therefore, the ball will move approximately 140.4225 meters.
To calculate the distance the ball will move, we can use the kinematic equation:
d = v₀t + (1/2)at²
Where:
d = distance
v₀ = initial velocity (which is 0 since the ball starts from rest)
t = time
a = acceleration
In this case, the initial velocity (v₀) is 0, the acceleration (a) is 4.5 m/s², and the time (t) is 7.9 s.
Plugging these values into the equation, we get:
d = 0(7.9) + (1/2)(4.5)(7.9)²
= 0 + (1/2)(4.5)(62.41)
= 0 + (2.25)(62.41)
= 140.42 m
Therefore, the ball will move approximately 140.42 meters down the hill.