Question

Question 5:- 5. When a car driver travelling at a speed of 20 m/s applies brakes and brings the car to rest in 20 s, then the retardation will be: a) + 2 m/s2 b) - 2 m/s2 c) +1.5 m/s2 d) + 1 m/s2​

Answer 1

Provided that:

• Initial velocity = 20 mps
• Final velocity = 0 mps
• Time = 20 seconds

To calculate:

• The retardation

Solution:

• The retardation = -1 m/s²

Using concept(s):

• We can solve this question by using either acceleration formula or the first equation of motion.

• Choice may vary!

Using formula(s):

• Acceleration formula:

• a = (v-u)/t

• First equation of motion:

• v = u + at

Where, a denotes acceleration, v denotes final velocity, u denotes initial velocity, t denotes time.

Required solution:

~ Finding retardation by using acceleration formula firstly!

→ a = (v-u)/t

→ a = (0-20)/20

→ a = -20/20

→ a = -1 mps sq.

→ Retardation = -1 mps sq.

~ Now by using first equation of motion let us calculate the acceleration!

→ v = u + at

→ 0 = 20 + a(20)

→ 0 = 20 + 20a

→ 0 - 20 = 20a

→ -20 = 20a

→ -20/20 = a

→ -1 = a

→ a = -1 mps sq.

→ Retardation = -1 mps sq.

Additional information:

$\begin{gathered}\boxed{\begin{array}{c}\\ {\pmb{\sf{Three \: equation \: of \: motion}}} \\ \\ \sf \star \: v \: = u \: + at \\ \\ \sf \star \: s \: = ut + \: \dfrac{1}{2} \: at^2 \\ \\ \sf \star \: v^2 - u^2 \: = 2as\end{array}}\end{gathered}$

Answer 2

Answer:

♤Given :-

• When a car driver travelling at a speed of 20 m/s applies breaks and brings the car to rest in 20s.

□To prove :-

• What is the retardation.

□Explanation :-

• Here by using first equation of motion formula we get that,
• $v = u + at$

♧Now applying the values we get that ,

• $0 = 20 + a(20)$
• $0-20= 20 a$
• $-20 = 20 a$
• $a = \frac{ - 20}{ \: \: \: 20}$
• $a = - 1mps \: sq.$

♧Therefore , the retardation is -1mps sq.

◇Used formula :-

• First equations of motion formula we used here to get the answer.

□Hope it helps u mate .

□Thank you .