Initial velocity u = 30 m/s
As it stops then final velocity v = 0 m/s
Time taken t = 2.0 s
We know,
v-u = at
=> 0-30= 2a
=> a =
-ve sign shows velocity is decreasing.
Initial velocity u = 60 m/s
As it stops so final velocity v = 0 m/s
Time taken t = 2 seconds
We know
v-u = at
Also
From (1), we have
After putting given values, we have
=> s = 120-60 = 60 m
Initial velocity u = 30 m/s
As it stop so final velocity v = 0 m/s
Time t = 2 s
Distance covered = s
We know,
s =
=> s = 30 m
Initial velocity u = 0 m/s
final velocity = v
Time t = 2 s
Acceleration a = 1 m/s2
We know,
v = u + at
=> v = 2 m/s
A truck has a velocity of 2 m /s at time t=0. It accelerates at 2 m / on seeing police .What is its velocity in m/s at a time of 2 sec
Initial velocity u = 2 m/s
final velocity = v m/s
Time duration = final time - initial time = 2-0 = 2 s
acceleration a = 2 m/s2
We know,
v = u + at
=> v =
=> v = 6 m/s
Initial velocity u = 3 m/s
Acceleration a = 3 m/s2
Initial time
Final time
Time taken t = 2-0= 2 s
Final velocity v = ?
We know,
A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. Determine the time in seconds at which the stone reaches its maximum height. g =9.8 m /
Initial velocity u = 20.0 m/s
At maximum height it ll stop
So final velocity v = 0 m/s
Acceleration due to gravity g = 9.8 m/s2
Time taken to reach maximum height = t
We know
v = u + at
=>
[g is taken negative because it is in opposite direction of motion.]
A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. Determine the maximum height it travels in meters. g =9.8 m /
Initial velocity u = 20.0 m/s
At maximum height stone ll be stopped,
So final velocity v = 0 m/s
Acceleration due to gravity a = g =
-9.8 m/s2 (-ve Because it is in opposite direction of motion)
Let maximum height = s
We know,
=>
A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. Determine the velocity in m/sec when the stone returns to the height from which it was thrown. g =9.8 m /
we get
As the motion of the stone is downward, and the”+”sign was assigned for the upward motion,
we get for vC = −vA= −20 m/s.
A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The height of the building is 50.0 m. determine the velocity in m/sec when the stone hits the ground. g =9.8 m /
When the stone ll reach at the same point from where is was thrown it ll have same velocity but with opposite sign.
So initial velocity u = -20 m/s
Final velocity before hitting ground = v
Distance covered s = 50 m
Acceleration due to gravity a = 9.8 m/s2
We know
As this velocity is in opposite direction is initial velocity so sign ll be negative.
v = -37.1 m/s