Question

A car initially at rest starts moving with an acceleration of 5m/s^2 covers a distance of 25m. Find the time required to cover this distance​

Answer 1

Given :

Initial velocity of the car = zero.

Acceleration of the car = 5m/s².

Distance covered = 25m.

To find :

The time required to cover that distance.

Solution :

As we are provided with initial velocity, accerlation and distance we can use second equation of motion .i.e.,

$\boxed{ \sf s = ut + \dfrac{1}{2}at^2}$

here,

• s denotes distance covered
• u denotes initial velocity
• t denotes time
• a denotes acceleration

by substituting all the given values in the equation,

$\leadsto{ \sf s = ut + \dfrac{1}{2}at^2}$

$\leadsto{ \sf 25 = (0)t + \dfrac{1}{2}(5)t^2}$

$\leadsto{ \sf 50 = 5t^2}$

$\leadsto{ \sf t ^{2} = \dfrac{50}{5} }$

$\leadsto{ \sf {t}^{2} = 10 }$

$\leadsto{ \sf t = \sqrt{10} }$

$\leadsto \underline{ \boxed{ \sf t = 3.16 \: sec }}$

thus, the time taken by the car is 3.16 sec.

Know more :

Acceleration is measured as the rate of change of velocity.

• It is a vector quantity having both magnitude and direction
• SI unit of accerlation is metre per second square

Velocity is the rate of displacement of a moving object over time.

• Velocity is vector quantity having both magnitude and direction
• SI unit of velocity is metre per second

Answer 2

Given:

Initial velocity of the car = zero.

Acceleration of the car = 5m/s².

Distance covered = 25m.

To find:

Time required to cover that distance = ?

Solution:

As we are provided with initial velocity, accerlation and distance we can use second equation of motion .i.e.,

$\boxed{ \sf s = ut + \dfrac{1}{2}at^2}$

$\leadsto{ \sf s = ut + \dfrac{1}{2}at^2}$

$\leadsto{ \sf 25 = (0)t + \dfrac{1}{2}(5)t^2}$

$\leadsto{ \sf 50 = 5t^2}$

$\leadsto{ \sf t ^{2} = \dfrac{50}{5} }$

$\leadsto{ \sf {t}^{2} = 10 }$

$\leadsto{ \sf t = \sqrt{10} }$

$\leadsto \underline{ \boxed{ \sf t = 3.16 \: sec }}$

Your answer is 3.16 sec .

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