Question

A car accelerates uniformly from 5m s-1 to 15m s-1 taking 7.5 seconds. How far did it travel during this period.

Answer 1

Given :-

• Initial Velocity (u) = 5 m/s
• Final Velocity (v) = 15 m/s
• Time (t) = 7.5 seconds

To Find :-

• Distance (s) = ?

Solution :-

First of all we will find the acceleration of car :

→ Acceleration = Final Velocity - Initial Velocity ÷ Time

→ Acceleration = 15 - 5 ÷ 7.5

→ Acceleration = 10 ÷ 7.5

→ Acceleration = 1.3 m/s²

Now, by using Second equation of motion we can find distance travelled by the car :

→ s = ut + ½ at²

→ s = 5 × 7.5 + ½ × 1.3 × (7.5)²

→ s = 37.5 + ½ × 1.3 × 56.25

→ s = 37.5 + 1.3 × 28.125

→ s = 37.5 + 29.425

→ s = 66.925 m

Therefore,the distance travelled by the car is 66.925 m.

Answer 2

Question :-

A car accelerates uniformly from 5m s-1 to 15m s-1 taking 7.5 seconds. How far did it travel during this period.

Answer :-

✦ Distance travelled by the car is 66.925 m.

Given :-

A car accelerates uniformly from 5m s-1 to 15m s-1 .

So ,

(u) = 5 m/s

(v) = 15 m/s

Time Taken = 7.5 seconds

Have to Find :-

Distance travelled by the car .

Required Solution :-

So , Lets find out acceleration first :-

So As we alll know ,

A = v - u / t

" A " = Acceleration

" v " = Final Velocity

" u " = Initial Velocity

"T " = Time Taken

Now , substituting the values in the above formula

A = 15 - 5 / 7.5

= 10 / 7.5

Therefore , The Acceleration is 1.3 m/s²

In order to Find the distance travelled let's apply the Second equation of motion :-

{ s = ut + ½ at² }

Here ,

"s " = Distance travelled

" A " = Acceleration

" v " = Final Velocity

" u " = Initial Velocity

"T " = Time Taken

Substituting the values , we get :-

s = 5 × 7.5 + ½ × 1.3 × (7.5)²

↠ s = 37.5 + ½ × 1.3 × 56.25

↠ s = 37.5 + 1.3 × 28.125

↠ s = 37.5 + 29.425

↠ s = 66.925 m { Distance travelled }

Hence ,

The required answer is 66.925 m which is the distance travelled by the car.

More related information :-

✮ First equation of motion

❥ v = u + at.

v = final velocity.

u = initial velocity.

a = acceleration.

t = time taken.

Note :

first equation of motion also called velocity time equation.

✮ Second equation of motion

↠ s = ut + ½ at².

u = initial velocity.

a = acceleration.

t = time taken.

s = distance covered.

Note :-

Second equation of motion also called positive time equation.

✮ Third equation of motion

❥ v² - u² = 2as

v = final velocity.

u = initial velocity.

a = acceleration.

Note :-

Third equation of motion also called velocity positive equation.

❥ First law of motion

↠ first law of motion states that an object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.

❥ Second law of motion

↠ second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.

❥ Third law of motion

↠ third law of motion states that for every action there is an equal and opposite reaction and they act on two different bodies.