Question

a toy car was moving with a speed of 20m/s.It speed became 50m/s after traversing a distance of 1m. calculate the acceleration of the car.​

Answer 1

Given:

Initial velocity of toy car,u= 20 m/s

Final velocity of toy car,v= 50 m/s

Distance covered by toy car,s= 1 m

To Find:

Acceleration of the car

Solution:

We know that,

• According to third equation of motion for constant acceleration,

v²- u² = 2as

where,

v is final velocity

u is initial velocity

a is acceleration

s is displacement

━━━━━━━━━━━━━━━━━━━━━

Let the acceleration of toy car be a

So, on applying third equation of motion on toy car, we get

➩ v²-u²= 2as

➩ (50)²-(20)²= 2a(1)

➩ 2500-400= 2a

➩ 2100= 2a

➩ 2a=2100

➩ a= 2100/2

➩ a= 1050 m/s²

Hence, the acceleration of toy car is 1050 m/s².

Answer 2

$\huge{\mathcal{\purple{G}\green{i}\pink{v}\blue{e}\purple{n}\green{:}\pink{-}}}$

• Initial velocity (u) of car = 20m/s

• Final velocity (v) of car = 50m/s

• Distance covered by car (s) = 1 m.

$\huge {\underline{\pink{To Find:-}}}$

• Acceleration of the car.

$\huge {\underline{\pink{Solution:-}}}$

According to the question we have given the initial and Final velocity of the car and the distance travelled by car. So, we have to find the acceleration of the car.

So, By using 3rd equation of motion

➲ v² = u² +2as

Putting all the values which is given above

⇒ 50² = 20² + 2×a ×1

⇒ 2500 = 400 + 2× a

⇒ 2500-400 = 2× a

⇒ 2100 = 2×a

⇒a = 2100/2

⇒ a = 1050 m/s²

∴ The acceleration of the car is 1050 m/s².

$\huge {\underline{\pink{Additional Information :-}}}$

• The rate of change of velocity at per unit time is called Acceleration.

• SI unit of Acceleration is m/s²

• It is vector Quantity.