Initial distance x1 = 1000m
Initial time t1= 0 sec
Final distance x2 = 900 m
Final time t2 = 10 sec
Average velocity
=
Total distance covered s = 1000 m
Total time taken t = 100-0 = 100 s
Average speed
Total distance covered s = 1000 m
Total time taken t = 100-0 = 100 s
Average speed
Average acceleration is the change in velocity divided by an elapsed time.
Initial velocity u = 0 m/s
Final velocity v = 20 m/s
Time elapsed = 10-0 = 10 s
Average acceleration
Average velocity can be defined as the displacement divided by the time.
Total displacement d = 18 m
Time taken t = 1 s
average velocity
Let the distance traveled by the bus when man catch the bus be S meter distance traveled by the man when he catch the bus S'=S+48 meter
Given :
Initial velocity of bus u=0
Initial velocity of man u'=10m/sec Acceleration of Bus=1 m/s2
Accelerationof Man=0
We will use formula for displacement
By substituting the values in above equation we get,
Displacement of Bus
Displacement of Man
By substituting the value of S from eq(1) in eq(2)we get
=> t = 8, 12
Taking minimum value t = 8 sec
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.]
Instantaneous speed is the slope of the line at that point.
According to given image
Slope of line at x2 =
= 122.7
Initial velocity u = 60 m/s
As it stops then final velocity v = 0 m/s
Time taken t = 2.0 s
We know,
v-u = at
=> 0-60= 2a
=> a =
-ve sign shows velocity is decreasing.