Question

if a car have running at a speed of 10m/s and acceleration -2m/s^2 and a time of 6 sec find the distance covered by car​

Answer 1

Answer:

• 24 metres is the required answer .

Explanation:

According to the Question

It is given that,

• Initial velocity,u = 10m/s
• Acceleration ,a = -2m/s²
• Time taken ,t = 6s

we have to calculate the distance covered by the car .

we use here the kinematics equation of motion

• s = ut + ½ at²

On substituting the value we get

↠ s = 10 × 6 + ½ × (-2) × 6²

↠ s = 60 - 6²

↠ s = 60 - 36

↠s = 24m

• Hence, the distance covered by the car is 24 metres.

Answer 2

Answer:

Given :-

• A car have running at a speed of 10 m/s and acceleration - 2 m/s² and a time of 6 seconds.

To Find :-

• What is the distance covered by car.

Formula Used :-

$\clubsuit$ Second Equation Of Motion Formula :

$\bigstar \: \: \sf\boxed{\bold{\pink{s =\: ut + \dfrac{1}{2} at^2}}}\: \: \: \bigstar$

where,

• s = Distance Covered
• u = Initial Velocity
• t = Time Taken
• a = Acceleration

Solution :-

Given :

• Initial Velocity = 10 m/s
• Acceleration = - 2 m/s²
• Time Taken = 6 seconds

According to the question by using the formula we get,

$\implies \bf s =\: ut + \dfrac{1}{2} at^2$

$\implies \sf s =\: (10)(6) + \dfrac{1}{2} \times (- 2)(6)^2$

$\implies \sf s =\: (10 \times 6) + \dfrac{1}{2} \times (- 2) \times (6 \times 6)$

$\implies \sf s =\: 60 + \dfrac{1}{2} \times (- 2) \times 36$

$\implies \sf s =\: 60 + \dfrac{1}{2} \times (- 72)$

$\implies \sf s =\: 60 + \dfrac{1 \times (- 72)}{2}$

$\implies \sf s =\: 60 + \dfrac{- 72}{2}$

$\implies \sf s =\: 60 + (- 36)$

$\implies \sf s =\: 60 - 36$

$\implies \sf\bold{\red{s =\: 24\: m}}$

$\small \sf\bold{\purple{\underline{\therefore\: The\: distance\: covered\: by\: car\: is\: 24\: m\: .}}}$