Not really.
For an object which accelerates from rest, the variation of speed with time is given by:
final velocity=initial velocity + acceleration * time
= 0 + 10 m/s² * 2 s
=20 m/s
(note how the units cancel to give m/s)
An object dropped from a cliff falls with a constant acceleration of 10m/s2. Find its speed 2s after it was dropped?
Solution:Given ,accelaration =10m/s
time=2s
speed=?
therfore,speed=distance travelled\total time taken
s=10\2=5(ans) . is it ok.
8 answers
We will apply v=u+at . 0+2×10= 20m/s
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As from the kinematics equation
V=u+at
So, u = 0
a = 10m/s2
t = 2 s
V = 0 + 10 * 2 = 20 m/s
V=u+at
So, u = 0
a = 10m/s2
t = 2 s
V = 0 + 10 * 2 = 20 m/s
Given,
a = 2 m/s^2
t = 10 s
u =0(Initial speed)
v = ? (Final speed)
By the equation v = u + at
v = 0 + 2*10
= 0 + 20
= 20 m/s
Therefore the speed 2 sec after it was dropped is 20 m/s
a = 2 m/s^2
t = 10 s
u =0(Initial speed)
v = ? (Final speed)
By the equation v = u + at
v = 0 + 2*10
= 0 + 20
= 20 m/s
Therefore the speed 2 sec after it was dropped is 20 m/s
2 answers