The displacement (in metre) of a particle moving along x-axis is given by x = 18t + 5t2 : calculate (i) the instantaneous velocity at t=4 sec (ii) average velocity between t=4 sec and t=5sec. (iii) instantaneous acceleration.
Answers
Answered by
140
instantaneous velocity v= dx/dt=18+10t
at t=4
v = 18+10×4=58 m/s
a= dv/dt =10 m/ s sq.
at t=5 v= 18+10×5=68 m/s
so avg velocity = (68-58)/5-4=10 m/s
pls mark it as brainliest.
at t=4
v = 18+10×4=58 m/s
a= dv/dt =10 m/ s sq.
at t=5 v= 18+10×5=68 m/s
so avg velocity = (68-58)/5-4=10 m/s
pls mark it as brainliest.
Answered by
94
Hey there!
Find out ur answer below!
1.)ds/dt = d/dt * (18t +5t^2)
☞ V = 18+ 10t
3.) dv/dt = d/dt* (18+10t)
=0 +10
☞ a = 10 m/s^2
2.) Average velocity =
at t = 4
x1 = 5(4)^2+ 18(4) = 152m
and at t = 5s
x2 = 5(5)^2+ 18(5) = 215m
so,
total distance travelled
dx = (x2 - x1)
total time taken
dt = 1s
so, average velocity
vav = dx/dt = (215-152) / 1
thus,
vav = 63 m/s
# hope it helps!
Find out ur answer below!
1.)ds/dt = d/dt * (18t +5t^2)
☞ V = 18+ 10t
3.) dv/dt = d/dt* (18+10t)
=0 +10
☞ a = 10 m/s^2
2.) Average velocity =
at t = 4
x1 = 5(4)^2+ 18(4) = 152m
and at t = 5s
x2 = 5(5)^2+ 18(5) = 215m
so,
total distance travelled
dx = (x2 - x1)
total time taken
dt = 1s
so, average velocity
vav = dx/dt = (215-152) / 1
thus,
vav = 63 m/s
# hope it helps!
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