Question

A block is moving with a velocity of 22 m/s. A constant
force is applied on the block in the direction opposite to
the direction of velocity. After 9 seconds, the velocity is
reduced to 13 m/s.
After this, the direction of the same force is reversed. It is
now applied in the direction of the velocity for 7 seconds.
What will be the velocity of the block after 7 seconds?
10 m/s​

Answer 1

we use first equation of motion

v-u=at

13-22=a*9

a=-1 metre per second

for another interval

v-13=1*7

v=20 m/s

Answer 2

Answer:

20 m/s is the final velocity.

Step-by-step explanation:

Initial velocity of the block (u) = 22 m/s

Final velocity of the block (v) = 13 m/s

Time take (t) = 9 sec

Applying the first equation of motion,

v - u = at

13 - 22 = a X 9

∴ a = - 1 m/s  ( Negative sign due to the force being applied to the opposite direction of velocity)

Now, initial velocity of the block (u) = 13 m/s

Time taken (t)  = 7sec

Applying the same equation as above,

v - 13 = 1 X 7   ( a is positive because force is in the same direction as the velocity)

∴ v = 20 m/s