Question

Displacement s is related with time t in the equation
$s = 5 {t}^{2} + 4t + 3$
The acceleration is equal to

Answer 1

The accekeration is equal to the Double derivative of the function s with reapect to time t...

s = 5 t^2 + 4t +3
on differentiating both sides with respect to t gives
ds / dt =( 10t + 4 ) m/s

Now
ds / dt = Change in displacement / Change in time
= velocity = v
so, ds / dt = v
Now ds / dt = v = 10t + 3
= > v = 10t + 3
On differentiting above equation with respect to t gives
dv / dt = 10
Here
dv / dt = Rate of chage of velocity = Acceleration = a
so , a = 10 m/ s^2

Answer 2

sorryyyy....I don't know the answer....