Question

Car acilaretes uniformly from 52 km/h to 90 km/h in 5 seconds calculate the acelaration and distance traveled by car at that time

Answer 1

Solution :—

Given :-

• Initial Velocity (u) = 52 km/hr = 14.4 m/s
• Final Velocity (v) = 90 km/hr = 25 m/s
• Time Taken (t) = 5 s

To Find :-

• Acceleration (a)
• Distance Covered (s)

Calculating Acceleration

Formula Used :-

• v = u + at

⟹ 25 = 14.4 + a × 5

⟹ 25 = 14.4 + 5a

⟹ 25 - 14.4 = 5a

⟹ 10.6 = 5a

⟹ a = 10.6/5

⟹ a = 2.12

Thus, acceleration is 2.12 m/s²

Calculating Distance Covered

Formula Used :-

• v² - u² = 2as

⟹ (25)² - (14.4)² = 2 × 2.12 × s

⟹ 625 - 207.36 = 4.24 s

⟹ 407.64 = 4.24s

⟹ s = 407.64/4.24

⟹ 98.5

Thus, Distance Covered is 9.85 m

Hence, Solved ♡

Answer 2

Answer :

›»› The Acceleration of car = 2.12 m/s²

›»› The Distance travelled by car = 98.5 m

Given :

• Initial Velocity of car = 52 km/hr
• Final Velocity of car = 90 km/hr
• Time Taken by car = 5 sec

To Find :

• Acceleration of car = ?
• Distance travelled by car = ?

Required Solution :

Here in this question we have to find Acceleration of car and Distance travelled by car So, firstly we have to convert the SI unit of Initial velocity and Final velocity of car km/hr to m/s, after all this we will find Acceleration of car and Distance travelled by car on the basis of conditions given above$.$

⌯ Initial velocity = 52 km/hr

→ Initial velocity = 52 × 5/18

→ Initial velocity = 2.88 × 5

→ Initial velocity = 14.4 m/s

⌯ Final velocity = 90 km/hr

→ Final velocity = 90 × 5/18

→ Final velocity = 5 × 5

→ Final velocity = 25 m/s

Now, we have Initial velocity and final velocity and Time taken by car,

• Initial velocity of car = 14.4 m/s
• Final velocity of car = 25 m/s
• Time taken by car = 5 sec

And we need to find Acceleration of car.

From first equation of motion

⇛ v = u + at

⇛ 25 = 14.4 + a × 5

⇛ 25 = 14.4 + 5a

⇛ 25 - 14.4 = 5a

⇛ 10.6 = 5a

⇛ 10.6/5 = a

⇛ 2.12 = a

⇛ a = 2.12 m/s²

║Hence, the Acceleration of car is 2.12 m/s².║

Distance travelled by car

From second equation of motion

⇛ s = ut + ½ at²

⇛ s = 14.4 × 5 + ½ × 2.12 × 5²

⇛ s = 72 + ½ × 2.12 × 25

⇛ s = 72 + ½ × 53

⇛ s = 72 + 1 × 26.5

⇛ s = 72 + 26.5

⇛ s = 98.5 m

║Hence, the Distance travelled by car is 98.5 m.║