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. Average velocity between t=2second and t= 3second
Answers
I) instantaneous velocity at (t=2s) = 38 m/s
II) instantaneous acceleration =
10 m/s²
The average velocity between t=2 seconds and t = 3 seconds = 43 m/s.
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
Average velocity = total displacement / time taken
At t = 2 , x = 56
At t = 3 , x = 99
Total displacement = 99 - 56 = 43m
Time taken = 1 seconds
Average velocity = 43/1 = 43 m/s
Average velocity between t=2 seconds and t = 3 seconds is equal to 43 m/s.
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