numericals of Exercise -2(C)Q1-Q6 of chapter-2 based on equation of motion of class-9

 1. A body starts from rest with a uniform acceleration 2 metre per second square. Find the distance covered by the body in 2 second?

Ans. Starts from rest it means initial velocity is zero. So, 

u=0m/s²

a=2m/s²

t=2s

s=?

S=ut + 1/2 at²

S=0×2s + 1/2 × 2m/s² × 2s × 2s

s= 0 + 4m

s= 4m.

2.a body starts with an initial velocity of 10 metre per second and acceleration 5 metre per second square. Find the distance covered by it in 5 seconds?

Ans. Initial velocity e of 10 metre per second.

u= 10m/s

a=5m/s²

t=5s

S=?

s=ut + 1/2 at²

s= 10ms-¹ × 5s + 1/2 × 5ms-² ×5s × 5s

s= 50m + 62.5m

s= 112.5m

3. A vehicle is accelerating on a straight road. Its velocity at any instant is 30 kilometre per hour, after 2 seconds, it is 33.6 kilometre per hour and after further 2 seconds, it is 37.2 kilometre per hour. Find the acceleration of vehicle in metre per second square. Is deceleration uniform?

Ans.u1 = 30kmh-¹

              = 30×1kmh-¹

              = 30× 5/18 ms-¹

              = 5×5/3 ms-¹

              = 25/3ms-¹

              =8.33ms-¹

t1= 2s

u2 = 33.6 kmh-¹

      =33.6×1kmh-¹

      =33.6× 5/18 ms-¹

     =5.6 × 5/3 ms-¹

     =9.33ms-¹

t2= 2s

u3= 37.2kmh-¹

    =37.2×1kmh-¹

    =37.2× 5/18 ms-¹

    =6.2×5/3 ms-¹

    =10.33ms-¹

a= (v - u )/t

Case-1,

When velocity changes from u1 to u2  in 2 seconds and acceleration be a1.

a1= (u2- u1)/ t1

a1 = (9.33ms-¹ - 8.33ms-¹)/2s

a1= 1ms-¹ /2 s

a1= 1/2 ms-²

a1= 0.5 ms-²

Case-2,

When velocity changes from u2 to u3 in 2s,then acceleration be a2.

a2= (u3 - u2 )/t2

a2= (10.33ms-¹ - 9.33ms-¹)/2s

a2= 1ms-¹ / 2s

a2= 1/2 ms-²

a2= 0.5 ms-²

a1= a2 = 0.5 ms-²

So in both the cases acceleration remains same as 0.5 metre per second square. So we can say that acceleration is uniform.

4. A body initially at rest, thoughts moving with a constant acceleration 2 metre per second Square. Calculate: a) the velocity acquired b) the distance travelled in 5 seconds.

Ans. Body is initially at rest it means initial velocity is zero.

u=0ms-¹

a= 2ms-²

t=5s

v=?

s=?

v=u+at

v= 0 + 2ms-² × 5s

v= 0+ 10ms-¹

v=10ms-¹

s=ut + 1/2 at²

s=0×5s + 1/2 × 2ms-² × (5s)²

s= 0 + 25m

s=25m.

5.a bullet initially moving with a velocity 20 metre per second strikes a target and comes to rest after penetrating a distance 10 centimetre in the target. Calculate the retardation caused by the target.

Ans. Initial velocity is 20 metre per second. Since the body comes to rest after penetrating a distance of 10 centimetre in the target so its final velocity is zero.

u=20ms-¹

v=0

s= 10cm

  =10×1cm

   =10× 1/100 m

  =0.1m

a=?

v² - u² = 2as

2as=v² - u²

a= (v² - u²)/2s

a={o² - (20ms-¹)²}/2×0.1m

a= - 400m²s-²/0.2m

a= - 2000ms-²

Retardation= -a

 Retardation= - (- 2000ms-²)

                       = + 2000ms-²

6.a train moving with a velocity of 20 metre per second is brought to rest by applying brakes in 5 seconds. Calculate the retardation?

Ans. Initial velocity,

u= 20m/s

Body is brought to rest so its final velocity will become zero.

v=0m/s

t=5s

a=?

v=u+at

v- u = at

(v- u)/t =a

a= (v- u)/t

now putting the values of initial velocity final velocity and time in the equation we get,

a=( 0 - 20ms-¹)/5s

a= -20ms-¹ / 5 s

a= - 4 ms-².

Retardation= - a

Retardation= - (-4ms-²)

                    =+ 4 ms-².