Question

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Answer 1

Explanation:

1. Given:

Initial velocity = u = 0 m/s

Time = t = 10 seconds

Distance = s = 60 metres

Using the second equation of motion :

S=ut+ 1/2 at^2

60=0×10+1/2×a×100

60=50a

a= 6/5

a = 1.2 m/s^2

The acceleration is equal to 1.2 m/s^2

2.Given:

Initial velocity = u = 60m/s

Distance = s = 100 metres

Time = t = 3 seconds

S=ut+1/2at^2

100=60×3+1/2a×9

100=180+4.5a

-80 = 4.5a

a = - 17.777

V=u+at

V = 60-17.77×3

V = 8...

So the bus will have some velocity and the boy will be hence hit

3. Given:

Initial velocity = u = 0m/s

Final velocity = v = 50 m/s

Time = t = 5 seconds

Acceleration = (v-u) /t

Acceleration = 50-0/5

Acceleration = 10 m/s^2

Distance can be found using second equation of motion :

S =ut+1/2 at^2

S = 0×5+1/2×10×25

S = 125 metres

4. Given:

For ball going up

Initial velocity = u = 20 m/s

Acceleration = - 10 m/s^2

V = 0 m/s

-10 = 0-20/t

t = 2 seconds for going up

Distance = 20 ×2+1/2×-10×4

Distance = 20m

Coming down:

Initial velocity = u = 0 m/s

Acceleration = 10 m/s^2

Distance = 20m

S=ut+1/2at^2

20=0×t+1/2×10×t^2

20=5t^2

t = 2 seconds

Total time = 2+2 = 4 seconds

Answer 2

1) We have given that, car starting from rest moves for 10 seconds & covers a distance of 60 m.

We have to find acceleration & final velocity.

☢ According to 2nd equation of motion :

● s = ut + ½ at²

⇒ 60 = 0 × 10 + ½ × a × 10²

⇒ 60 = 0 + ½ × a × 100

⇒ 60 = 50a

⇒ a = 60/50

⇒ a = 1.2 m/s²

∴ Acceleration of car = 1.2 m/s²

☢ Now using 1st equation of motion :

● v = u + at

⇒ v = 0 + 1.2 × 10

⇒ v = 0 + 12

⇒ v = 12 m/s

∴ Final velocity of car = 12 m/s

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2) We have, initial speed of bus is 60 m/s,it stops 3 seconds after applying brakes. Child is at 100 m ahead of the bus.

We have to find if the child is saved or not.

Here, u = 60 m/s

Finsl velocity (v) = 0 m/s [As it stops]

Time = 3 seconds.

☢ Using 1st equation of motion :

● v = u + at

⇒ 0 = 60 + a × 3

⇒ 3a = -60

⇒ a = -60/3

⇒ a = -20 m/s²

☢ Using 3rd equation of motion :

● v² - u² = 2as

⇒ 0² - 60² = 2s × (-20)

⇒ 0 - 3600 = -40s

⇒ -40s = -3600

⇒ s = -3600/-40

⇒ s = 90 m

∴ Distance covered by bus = 90 m

As the bus covers 90 m as distance and child is at 100 m ahead.

So we can say that,

∴ Child will be saved from being hit by bus.

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3) We have car starts from rest attains velocity of 50 m/s in 5 seconds.

We have to find distance covered & acceleration.

Here, u = 0 m/s

v = 50 m/s

t = 5 seconds.

☢ Using the equation :

● a = (v - u)/t

⇒ a = (50 - 0)/5

⇒ a = 50/5

⇒ a = 10 m/s²

∴ Acceleration of car = 10 m/s²

☢ Using 2nd equation of motion :

● s = ut + ½ at²

⇒ s = 0 × 5 + ½ × 10 × 5²

⇒ s = 0 + 5 × 25

⇒ s = 125 m

∴ Distance covered by car = 125 m

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4) Ball thrown vertically upwards with velocity of 20 m/s

We have to find time taken by ball to reach ground.

At first, while going upwards,

Initial velocity (u) = 20 m/s

Final velocity (v) = 0 m/s

Let the acceleration (g) be = -10 m/s²

☢ Using 1st equation of motion :

● v = u + at

⇒ 0 = 20 + (-10)t

⇒ 20 - 10t = 0

⇒ 20 = 10t

⇒ t = 20/10

⇒ t = 2 seconds.

Again, while coming back to ground, initial velocity = 0 m/s & final velocity = 20 m/s

Gravitational acceleration (g) = 10 m/s²

☢ Using 1st equation of motion :

● v = u + at

⇒ 20 = 0 + (10)t

⇒ 10t = 20

⇒ t = 20/10

⇒ t = 2 seconds.

Hence, total time taken :

⇒ Total time = 2 + 2

⇒ Total time = 4 seconds.

∴ Total time taken by ball to reach ground = 4 seconds.