Question

A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car and the distance traveled ​

Answer 1

Given:

• Initial velocity = 18.5 m/s
• Final velocity = 46.1 m/s
• Time taken = 2.47 s

To find:

• Acceleration
• Distance travelled

Solution:

Firstly finding acceleration

Using ,

a = v - u/t

Where

• a → acceleration = ?
• v → final velocity = 46.1 m/s
• u → Initial velocity = 18.5 m/s
• t → time taken = 2.47 s

Substituting the values we have

→ a = 46.1 - 18.5/2.47

→ a = 27.6/2.47

→ a = 11.17 m/s²

________________________

Now , finding the distance travelled by using the kinematic equation

S = ut + 1/2 at²

Where

• s → distance = ?
• u → initial velocity = 18.5 m/s
• a → acceleration = 11.17 m/s²
• t → time taken = 2.47s

Substituting the values we have

→ s = 18.5(2.47) + 1/2 × (11.17)(2.47)²

→ s = 45.7 + 34.1

→ s = 79.8 m

★ Answers :

• Acceleration = 11.17 m/s²
• Distance = 79.8 m

Answer 2

Explanation:

${\red{\underline{\underline{\bold{Given:-}}}}}$

• Initial velocity = 18.5 m/s

• Final velocity = 46.1m/s

• Time = 2.47 sec

${\blue{\underline{\underline{\bold{To\:Find:-}}}}}$

• Acceleration of the car

• Distance traveled by the car

${\green{\underline{\underline{\bold{Solution:-}}}}}$

According to first equation of motion

$\large \boxed{ \pink{\rightarrow\rm v = u + at }}$

Here:-

• v = final velocity

• u = initial velocity

• a = acceleration

• t = time

Substituting the values:-

$\rightarrow\rm 46.1 = 18.5 + a(2.47)$

$\rightarrow\rm 46.1 - 18.5 = 2.47a$

$\rightarrow\rm 27.6 = 2.47a$

$\rightarrow\rm \dfrac{27.6}{2.47} = a$

$\rightarrow\rm a = 11.17 m/{s}^{2}$

According to third equation of motion

$\large \boxed{ \pink{\rightarrow\rm {v}^{2} - {u}^{2} = 2as }}$

Here:-

• s = distance covered

Substituting the values

$\rightarrow\rm {46.1}^{2} - {18.5}^{2} = 2(11.17)(s )$

$\rightarrow\rm 2125.21 - 342.25 = 22.34s$

$\rightarrow\rm 1782.96 = 22.34s$

$\rightarrow\rm \dfrac{1782.96}{11.17} = s$

$\rightarrow\rm s = 79.8$

Therefore:-

$\boxed{\rm \purple{ \rightarrow Acceleration \: = 11.17m /{s}^{2} }}$

$\boxed{\rm \purple{ \rightarrow Distance \: = 79.8m }}$