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.
We will apply v=u+at . 0+2×10= 20m/s
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)
As from the kinematics equation
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
The formula you used to calculate the speed of the object is incorrect. To find the speed of an object, you need to multiply the acceleration by the time elapsed.
In this case, the acceleration is given as 10 m/s^2, and the time is given as 2 seconds. Thus, the formula to calculate the speed would be:
speed = acceleration × time
Plugging in the given values:
speed = 10 m/s^2 × 2 s = 20 m/s
So, the correct speed is 20 m/s, not 5 m/s.