Question

a ball rolls down a frictionless inclined plane with a uniform acceleration of 1 m/s^2 .if it's velocity at some instance of time is 10m/s, what will be its velocity 5 s later?

Answer 1

Answer:

a =

(v

(t

, so vf = a(t (where vi = 0 )

vf = 1.5 m

s

2

⋅3s = 4.5 m

s

Now we can use the freefall formula (or maybe we should call

it the “smooth acceleration from an initial velocity of zero” formula, d =

1

2

at

2

, to

calculate the distance: d =

1

2

&1.5m⋅s

−2

'&3s'

2 = 6.75m .

For the last part, we can use the acceleration definition, a =

(v

(t

, to find the time that

it takes for the velocity to reach 15 m/s: (t =

(v

a

=

15 m· s

−1

1.5 m· s

−2 = 10 s . Then the

distance covered in that time will be d =

1

2

at

2 =

1

2

&1.5m· s

−2

'&10 s'

2 = 75 m

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Answer 2

Answer:

15m/s

Step-by-step explanation:

acc = 1

velocity = 10

time = 5 sec

vf = vi + at

= 10 + 1×5

= 10 + 5

vf = 15m/s