Motion in A Straight Line - Online Test

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 = −302=−15 m/s2

-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

=>a=v−ut  .....(1)

Also

s=ut+12at2

From (1), we have

s=ut+12(v−ut)t2

=>s=ut+12(v−u)t

After putting given values, we have 

=>s=(60×2)+12(0−60)×2

=> 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 = 12(u+v)t

=>s=12(30+0)×2

=> s = 30 m

Initial velocity u = 0 m/s

final velocity = v 

Time t = 2 s

Acceleration a = 1 m/s​​​​​​2

We know,

v = u + at

=>v=0+1×2

=> v = 2 m/s

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/s​​​​​​2

We know,

v = u + at

=> v = 2+2×2

=> v = 6 m/s

Initial velocity u = 3 m/s

Acceleration a = 3 m/s​​​​​​2

Initial time t1=0 s

Final time t2=2 s

Time taken t = 2-0= 2 s 

Final velocity v = ?

We know,

v=u+at

=>v=3+2×3

=>v=3+6=9 m/s2

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/s​​​​​​2

Time taken to reach maximum height = t

We know

v = u + at

=> 0=20+(−9.8)t

=>t=−20−9.8=2.04 s [g is taken negative because it is in opposite direction of motion.]

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/s​​​​​​2 (-ve Because it is in opposite direction of motion)

Let maximum height = s

We know,

v2−u2=2as

=> 02−(20)2=2×(−9.8)s

=>−400=−19.6s

=>s=−400−19.6=20.4 m

(vC)2−(vA)2=2a(yC−yA)

With yC=yAwe get

(vC)2=(vA)2⟹vC=±vA

As the motion of the stone is downward, and the”+”sign was assigned for the upward motion,

we get for v​​​​​C = −vA= −20 m/s.

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/s​​​​​​2

We know

v2−u2=2as

=>v2−(−20)2=2×9.8×50

=>v2−400=980

=>v2=1380

=>v=1380−−−−√=±37.1

As this velocity is in opposite direction is initial velocity so sign ll be negative.

v = -37.1 m/s