Question

Question:-[ The brakes applied to a car produce an acceleration of 6m/s in the opposite direction to the motion.if the car takes 2s to stop after the application of brakes. Calculate the distance it's travel during time.

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Answer 1

Given:-

• Acceleration of the car (a) = -6m/s²

• Time taken (t) = 2s

• Final velocity (v) = 0m/s

To Find:-

• The distance covered by the car (s) .

Solution:-

Firstly we need find the initial velocity of the car

→ v = u+at

Substitute the value we get

→ 0 = u + 2×(-6)

→ -u = -12m/s

→ u = 12m/s

∴ The initial velocity of the car is 12m/s

Now using 2nd equation of motion

⟹ s = ut +1/2at²

Substitute the value we get

→ s = 12×2+ 1/2× (-6) ×2²

→ s = 24 + (-3) ×4

→ s = 24-12

→ s = 12m

∴ The distance covered by the car is 12metre.

Answer 2

Given :

• Acceleration (a) = 6 m/s
• Time (t) = 2 seconds
• Final velocity (v) = 0 m/s

To find :

• Distance (s)

According to the question,

$\star \boxed{ \sf \red{v = u + at}}$

Now according to the formula we will put the value

$\sf{ \: \: : \implies 0 = u + ( - 6) \times 2}$

$\sf{ \: \: : \implies0 = u + ( - 12) }$

$\sf{ \: \: : \implies0 = u - 12}$

$\sf{ \: \: : \implies u = 12 \: ms { }^{ - 1} } \: \orange\bigstar$

So,the initial velocity is 12 m/s...

Now we have to calculate the distance

$\star \boxed{ \sf \red{s = ut + \frac{1}{2} {at}^{2} }}$

According to the formula,

$\sf{ \: \: : \implies12 \times 2 + \frac{1}{2} \times ( - 6) \times {(2)}^{2} }$

$\sf{ \: \: : \implies 24+ \frac{1}{ \cancel2} \times ( - \cancel6) {}^{ \: 3} \times 4 }$

$\sf{ \: \: : \implies24 + ( - 12) }$

$\sf{ \: \: \: : \implies24 - 12}$

$\sf{ \: \: : \implies12 \: m} \: \orange \bigstar$

.°. The distance is 12 m...