Question

The displacement(in metres) of a particle moving along x-axis is given by x=18t+5t^2.Calculate: i)instantaneous velocity at t=2s ii)instantaneous acceleration.

Answer 1

using diffrentiation method
x = 18t + 5t^2
dx / dt = 18 + 10t
v = 18 + 10t 
t = 2 sec
v = 18 + 10 * 2 
= 18 + 20 
= 38 m / sec
t = 3 sec
v = 18 + 10 * 3
= 18 + 20
= 48 m / s
v(av) = 38 + 48/ 2 
= 43 m/swc
dv / dt = 10
a = 10 m/ s

Answer 2

Answer:

Given , x = 18t + 5t²

This equation is a relation between distance and time. For finding a relation between velocity and time , we will differentiate x w.r.t  t.

v = dx/dt

v = 10t + 18

Instantaneous velocity at t = 2 seconds = 10 × 2 + 18

= 38 m/s

For acceleration we will differentiate V w.r.t x.

acceleration = a = dv/dt = 10

Instantaneous acceleration = 10 m/s²