Question

Accelerating uniformly along a straight road at the rate of 1.5 m.s-2, a bus starts from rest and reaches a velocity of 15 m.s-1 in 6 seconds. Determine the distance travelled by the bus.

Answer 1

Correct question :-

Accelerating uniformly along a straight road at the rate of 2.5 ms⁻², a bus starts from rest and reaches a velocity of 15 ms⁻¹ in 6 seconds. Determine the distance travelled by the bus.

◘ Given ◘

► Accelerating uniformly along a straight road at the rate of 2.5 ms⁻², a bus starts from rest and reaches a velocity of 15 ms⁻¹ in 6 seconds.

• Acceleration, a = 2.5 m/s²
• Initial velocity, u = 0    [as it starts from rest]
• Final velocity, v = 15 m/s
• Time, t = 6 secs

◘ To Find ◘

The distance travelled (s) by the bus.

◘ Solution ◘

We know :

$\boxed{\bf{\dag\ v^2-u^2=2as}}$

Substituting the values :-

⟶ (15)² - (0)² = 2 × 2.5 × s

⟶ 225 - 0 = 5s

⟶ 5s = 225

⟶ s = 225/5

⟶ s = 45 m

________________

We know :

$\boxed{\bf{\dag\ s=ut+\frac{1}{2}at^2}}$

Substituting the values :-

⟶ s = 0(6) + 1/2 × 2.5 × (6)²

⟶ s = 0 + 1.25 × 36

⟶ s = 45 m

Therefore, the bus travels a distance of 45 m.

Answer 2

Question : -

A bus starts from rest and reaches a velocity of 15 m/s in 6 seconds. Determine the distance travelled by the bus ?

Answer : -

Given : -

A bus starts from rest and reaches a velocity of 15 m/s in 6 seconds.

Required to find : -

• Distance travelled by the bus ?

Equation used : -

➊ v² - u² = 2as

2. v = u + at

Here,

v = Final velocity

u = Initial velocity

a = acceleration

s = displacement

t = time taken

Solution : -

A bus starts from rest and reaches a velocity of 15 m/s in 6 seconds.

we need to find the distance travelled by the bus ?

So,

From the given data we can conclude that ;

• Initial velocity of the bus ( u ) = 0 m/s

• Final velocity of the bus ( v ) = 15 m/s

• Time ( t ) = 6 seconds

Now,

Let's find the acceleration of the bus ?

Using the equation of motion ;

i.e. v = u + at

↠ 15 = 0 + a x 6

↠15 = 0 + 6a

↠ 15 = 6a

↠ 6a = 15

↠ a = 15/6

↠ a = 2.5 m/s²

Hence,

• Acceleration of the bus ( a ) = 2.5 m/s²

Now,

Let's find the displacement of the bus ?

For this we can use 2 methods !

1st Method : -

Using the equation of motion ;

i.e. v² - u² = 2as

↠ ( 15 )² - ( 0 )² = 2 x 2.5 x s

↠ 225 - 0 = 2 x 2.5 x s

↠ 225 = 2 x 2.5 x s

↠ 225 = 5 x s

↠ 225 = 5s

↠ 5s = 225

↠ s = 225/5

↠ s = 45 meters

2nd Method : -

Using the equation of motion ;

i.e. s = ut + ½ at²

↠ s = 0 x 6 + ½ x 2.5 x 6 x 6

↠ s = 0 + ½ x 2.5 x 6 x 6

↠ s = ½ x 2.5 x 6 x 6

↠ s = 2.5 x 3 x 6

↠ s = 45 meters

Hence,

• Displacement of the bus ( s ) = 45 meters