Question

A Car accelerates uniformly from 5ms-1 to 13ms-1 in 4.0 seconds. what is the acceleration of the car?​

Answer 1

GiVeN :-

• Initial velocity, u = 5 m/s
• Final velocity, v = 13 m/s
• Time, t = 4 s

To FiNd :-

The acceleration(a) of the car.

SoLuTiOn :-

We know,

$\dag\ \sf{a=\dfrac{v-u}{t} }$

Putting the values :-

$\sf{\hookrightarrow a=\dfrac{13-5}{4} }\\\\ \sf{\hookrightarrow a=\dfrac{8}{4} }\\\\ \boxed{\boxed{\sf{\hookrightarrow a=2\ m/s^2}}}$

Therefore, the acceleration of the car is 2 m/s².

Answer 2

Given :-

• initial velocity of Car, u = 5 ms⁻¹
• final velocity of Car, v = 13 ms⁻¹
• time taken , t = 4.0 s

To find :-

• acceleration of the car, a = ?

Formula used :-

• First equation of motion

    v = u + at

( where v is the final velocity , u is the initial velocity , t is the time taken and a is the acceleration )

Solution :-

Using first equation of motion

⇰ v = u + at

( putting known values )

⇰ 13 = 5 + a ( 4 )

⇰ 4 a = 13 - 5

⇰ a = 2  ms⁻²     ( Ans. )

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Definition of acceleration

→ Acceleration is known as the rate of change of velocity .

→ Acceleration is a vector quantity that means it has both magnitude and direction.

→ Negative acceleration is also called deceleration.

Second equation of motion

→   s = u t + 1/2 a t²

( where s is the distance covered , u is the initial velocity , a is the acceleration and t is time )

Third equation of motion

→  2 a s = v² - u²

( where a is the acceleration , s is the distance covered , v is the final velocity and u is the initial velocity )

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