Question

15.
A car moving with a velocity of 50 km/hr is brought to rest in 45 secs. Find out the
retardation in m/sec
a) 0.288 m/sec
c) 0.358 m/sec
b) 0.308 m/sec?
d) 0.368 m/sec​

Answer 1

Answer:

$\large{\underline{\boxed{\sf Option \: (b) = 0.308 \: m/s^2 }}}$

Step-by-step explanation:

Given that ;

Initial velocity (u) = 50 km/h

Converting km/h into m/s.

⇒ u = 50 * (5/18)

⇒ u = 13.88 m/s

Final velocity (v) = 0 m/s

Time taken = 45 secs

Retardation (-a) =?

_______________________

We will use first equation of motion ;

⇒ v = u + at

⇒ 0 = 13.88 + a * 45

⇒ - 45a = 13.88

⇒ - a = $\sf \dfrac{13.88}{45}$

⇒ - a = 0.308 m/s²

_______________________

Hence, the retardation is 0.308 m/s².

Answer 2

☞ A car is moving with velocity 50 km/hr is brought to rest in 45 sec.

Here..

• Initial velocity (u) = 50 km/hr

=> $\dfrac{50\:\times\:5}{18}$

=> $\dfrac{125}{9}$ m/s

• Time (t) = 45 sec

• Final velocity (v) = 0

Final velocity is zero because car is brought at rest.

• We have to find retardation (-a).

______________________________

We know that

$\textbf{v\: = \:u\: + \:at}$

• Now put the given values in above equation.

=> 0 = $\dfrac{125}{9}$ + a (45)

=> 0 = 13.88 + 45a

=> - 13.88 = 45a

=> 45a = - 13.88

=> a = $\dfrac{-13.88}{45}$

=> a = - 0.308 m/s².

________________________________

Retardation is 0.308 m/s²

Option (b)

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