Question

The brakes applied to a car produce an acceleration of 6 m/s² in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.

Answer 1

Given :-

• Acceleration ( a ) is -6 m/s² [ negative sign is because it is in opposite direction of motion ]  
• Time ( t ) is 2 seconds  
• Final velocity ( v ) of car is 0 m/s  

To Find :-

• Distance ( s )  the car travels during this time.  

Solution :-

~Here, we’re given the acceleration , time , final velocity ( 0 as it comes to rest after applying brakes ) and we need to find the distance the car travels during this time. Firstly we’ll find the initial velocity ( u ) of the car by the first equation of motion and then we can easily find the distance by second equation of motion.

_____________

As we know that,

➲ First equation of motion : v = u + at

➲ Second equation of motion : s = ut + ½ × at²

Where,

• v is final velocity  
• a is acceleration  
• u is initial velocity  
• t is time  
• s is distance  

_____________

Finding the initial velocity :-

➟ v = u + at  

➟ 0 = u + ( -6  × 2 )  

➟ 0 = u + ( -12 )  

➟ 0 = u -12  

➟ u = 0 + 12

➟ u = 12 m/s  

Finding the distance :-

➟ s = ut + ½  × at²  

➟ s = ( 12 × 2 ) + ½  ×  ( -6 ) × ( 2 ) ²  

➟ s = 24 + ½ × ( -3 ) × ( 4 )  

➟ s = 24 + ( -12 )  

➟ s = 24 - 12  

➟ s = 12 m

Hence,  

• Distance covered by car during that time is 12m  

Answer 2

$\\ \underbrace{\color{green} \underline {\large \sf \: Question :-}}$

The brakes applied to a car produce an acceleration of 6 m/s² in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time.

$\\ \underbrace{\color{green} \underline {\large \sf \: Solution :-}}$

In the Question, It's is given that

★ Acceleration ( a ) = $\sf \: - 6m/s ^{2}$

★ Time ( t ) = 2s

★ Final velocity ( v ) = 0 m/s

At first ,

$\\ \: \: \: \: \sf \: \: \: formula :- \: \: \: \: \boxed{ \rm \: v = u + at \: \: }$

put given values in above formula ,

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: 0 = u - 6 \times 2$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: 0 = u - 12$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{ \pink{\frak {u= 12 \: m/s }}}$

Now , At second

$\\ \: \: \: \: \sf \: \: \: formula :- \: \: \: \: \boxed{ \rm \: s= ut+ \frac{1}{2} a {t}^{2} \: \: }$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: s = 12 \times 2 + \frac{1}{2} \times ( - 6) \times 4$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: s = 24 \times ( - 3) \times 4$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: s = 24 - 12$

$\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{ \pink{\frak {s= 12m}}}$

Answer :- 12 m Distance covered that time

$\\ \: \: \: \rm \red{Hope \: it's \: helpful }$