Question

A racing car has uniform acceleration of 4m/s2. How much distance will it cover in 1min after the start and what will be the final velocity of the car after 20sec??

Answer 1

SOLUTION :

Given :

▪ Initial velocity of car = zero

▪ Acceleration of car = 4m/s^2

To Find :

▪ Distance covered by car in 1min.

▪ Final velocity of car after 20s.

Concept :

✏ Since acceleration is constant throughout the entire journey, we can apply equations of kinematics directly..

✏ Acceleration is a vector quantity.

• First equation of kinematics

☞ v = u + at

• Third equation of kinematics

☞ S = ut + (1/2)at^2

Calculation :

✒ Distance travelled :

→ S = ut + (1/2)at^2

Putting given values, we get...

→ S = (1/2)×4×(60)^2 ๏[1min = 60s]๏

→ S = 2×3600

→ S = 7200m = 7.2km

✒ Final velocity :

→ v = u + at

→ v = 0 + 4(20)

→ v = 80mps

Answer 2

The velocity  is 80m/s.

The distance is 7200 m.

Explanation:

Given :

Initial velocity of car (u) = zero

Acceleration of car (a) = 4 m/s²

Time (t) = 20 seconds.

To Find :

The final velocity of the car & the distance.

We know that :

• Final Velocity :

$\bigstar\;\boxed{\bf a = v-\dfrac{u}{t}}}$

So,

$\implies \sf 4 = v - \dfrac{0}{20}\\ \\ \\\implies \sf 4 = \dfrac{v}{2}0\\ \\ \\\implies \sf 4 \times 20 = v\\ \\ \\\implies \sf 80 = v\\ \\ \\\implies \sf v = 80$

∴ The final velocity is 80 m/s.

$\rule{300}{1.5}$

• Distance Travelled :

We know that :

$\bigstar\;\boxed{\bf s = ut + \dfrac{1}{2}\;at^2}}$

So,

$\implies \sf 0\times60 + \dfrac{1}{2}\times4\times60\times60\\ \\ \\\implies \sf 2\times3600\\ \\ \\\implies \bf 7200$

∴ The distance travelled is 7200 m.