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A ball rolls down a hill with a constant acceleration of 3.0 m/s.
a. If it starts from rest, what is its speed at the end of 4.0 s?
b. How far did the ball move in that 4.0 s?

Respuesta :

molly4566
starting velocity (u) =0
acceleration (a) = 3
time (t) = 4
final velocity (v) =?

v=at+u
v=3(4)+0
v=12m/s2

distance (s) =?
s=0.5(u+v)(t)
s=0.5(0+12)3
s=18m

Final speed and Distance cover in 4 second is 12 m/s and 24 meter.

Given that;

Constant acceleration = 3.0 m/s

Find:

Final speed after 4 second

Distance cover in 4 second

Computation:

v = u + at

v = 0 + (3)(4)

v = 12 m/s

S = ut + (1/2)at²

S = (0)(4) + (1/2)(3)(4)²

S = (1/2)(3)(4)²

S = (1/2)(3)(16)

S = (3)(8)

S = 24 meter

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