Question

a car having initial velocity 8m/s moves with a constant acceleration of 8m/s^2 on a straight road .the ratio of distance travelled by car in first ,third and fifth second is

Answer 1

Given :-

• The initial velocity of car (u) = 8m/s
• The acceleration of car = 8m/s²

To Find :-

• The Ratio of distance travelled by car in 1s , 3s and 5s = ?

Solution :-

• To calculate the ratio of distance travelled by car at first we have to assume that the final velocity of car be v and distance travelled by car be s. Let the all final velocity be v_(1) , v_(2) and v_(3).

Calculate begins :-

• As per the 1st equation of motion :-

⇒ v = u + at

⇒v = 8 + 8 × 1

⇒v = 16m/s

⇒Final velocity (v_1) = 16m/s

• Calculation for second velocity :-

⇒ v = 8 + 8 × 3

⇒v = 8 + 24

⇒v = 32m/s

⇒Final velocity (v_2) = 32m/s

• Calculation for 3rd velocity :-

⇒v = 8 + 8 × 5

⇒v = 8 + 40

⇒v = 48m/s

⇒Final velocity (v_3) = 48m/s

• Now calculate the ratio of distance travelled by car by applying 3rd equation of motion :-

⇒ v² = u² + 2a s_(1)

⇒ 16² = 8² + 2×8 s_(1)

⇒256 = 64 + 16s_(1)

⇒16s_(1) = 256 - 64

⇒ 16s_(1) = 192

⇒ S_(1) = 12m

• Calculation for Distance _1

⇒ 32² = 8² + 2×8 s_(2)

⇒1024 = 64 + 16 s_(2)

⇒16s_(2) = 1024 - 64

⇒ 16s_(2) = 960

⇒ S_(2) = 60m

• Calculation for Distance_(3)

⇒ 48² = 8² + 2×8 s_(3)

⇒2304 = 64 + 16 s_(3)

⇒16s_(3) = 2304 - 64

⇒ 16s_(3) = 2240

⇒ S_(3) = 140m

• Now calculate it's ratio :-

⇒ Ratio ( in 1s : 3s : 5s)

⇒S_(1) : S_(2) : S_(3)

⇒ 12 : 60 : 140

⇒3 : 15 : 35

Hence,

• The required ratio = 3 : 15 : 35

Answer 2

Given :-

A car having initial velocity 8m/s moves with a constant acceleration of 8m/s^2 on a straight road

To Find :-

The ratio of distance travelled by car in first ,third and fifth second is

Solution :-

At first we need to find final velocity in each case

◼ I n C a s e 1

Time = 1 second

Initial velocity = 8 m/s

Acceleration = 8 m/s²

◼ I n C a s e 2

Time = 3 seconds

Initial velocity = 8 m/s

Acceleration = 8 m/s²

◼ I n C a s e 3

Time = 5 seconds

Initial velocity = 8 m/s

Acceleration = 8 m/s²

◼ I n C a s e 1

${\boxed{\frak{\red{\underline{\ast s = ut + 1/2 at^2\ast}}}}}$

Where

s = distance

u = initial velocity

a =  acceleration

t = time

s = 8(1) + 1/2 × 8 × (1)²

s = 8 + 4 × 1

s = 8 + 4

s = 12 m

◼ I n C a s e 2

${\boxed{\frak{\red{\underline{\ast s = ut + 1/2 at^2\ast}}}}}$

Where

s = distance

u = initial velocity

a =  acceleration

t = time

s = 8(3) + 1/2 × 8 × (3)²

s = 24 + 4 × 9

s = 24 + 36

s = 60 m

◼ I n C a s e 3

${\boxed{\frak{\red{\underline{\ast s = ut + 1/2 at^2\ast}}}}}$

Where

s = distance

u = initial velocity

a =  acceleration

t = time

s = 8(5) + 1/2 × 8 × (5)²

s = 40 + 4 × 25

s = 40 + 100

s = 140 m

Finding ratio

Ratio = S₁ : S₂ : S₃

Ratio = 12 : 60 : 140

Divide by 4

Ratio = 3 : 15 : 35