The displacement (in metre) of a particle moving along x - axis is given by = 2+, where A = 2m and B = 3m. Calculate (i) average velocity between t = 3s and t = 5s. (ii) instantaneous velocity at t = 5s and (iii) instantaneous acceleration
Answers
Explanation:
ANSWER
x=5t
2
+18t
(1)v=
dt
dx
=10t+18
att=2s,v=10×2+18=38m/s
(2)Average,velocity=
time
displacement
att=2,x=5(2)
2
+18×2=56m
t=3,x=5(3)
2
+18×3=99m
Averagevelocity=
3−2
99−56
=43 m/s
Answer:
average velocity =16m/s
instantaneous velocity =20m/s
instantaneous acceleration =4m/s^2
Explanation:
x=At^2+B
when t=3s,
x=2x(3)^2+3
= 2x9+3
=21m
when t=5s,
x=2x(5)^2+3
=2x25+3
=50+ 3
=53m
(I) average velocity =53-21/5-3
=32/2
=16m/s
(II) instantaneous velocity at t=5s
=dx/dt
=d(At^2+B)/dt
=d(2x2t+3x0)/dt
=4t+0
at t=5s,=4x5
=20m/s
(III) instantaneous acceleration
=dv/dt
=d(4t)/dt
= 4x1xt^1-1
=4x1
a=4m/s^2
from the formula,
if y=x^n
then, dy/dx=nx^n-1