Question

The position of stone dropped from a cliff is given by:

= 5
²
Where x is in meters and t is in seconds. What is the acceleration of the

stone at t = 2s?​

Answer 1

Answer:

From equation of motion

we know that,

dx/dt =v.

so x= 5t^2

and dx/dt=10t which is equal to “v” but at t=2 second v=10*2 which is equal to “20”

it means at t=2 v=20 .

OR

you can solve from this way,

at t = 2 sec , x=5*2^2=20

and initial velocity is zero

so, v^2=u^2+2as

v=(2as)^1/2

v=20

Explanation:

. Another ans is there

By Newton’s law, displacement s = u t + 1/2 a t^2. Since there is no t term in the given expression, we infer initial velocity u = 0. Also we have 1/2 a = 5, and hence a = 10.

Now velocity v = u + at. Putting u =0 and a = 10, t = 2, v = 20 units. (like 20 m/s)

Answer 2

Answer:

a=10m/s

Explanation:

v=dx/dt

 =d(5t^2)/dt

v=10t    -(i)

a=dv/dt

 =d(10t)/dt     .: from (i)

a=10m/s