Question

A car moving with uniform velocity 60 km/hr retards in 10 second and reduces to 45 km/hr .find the retardation. ​

Answer 1

Provided that:

• Initial velocity = 60 kmph
• Final velocity = 45 kmph
• Time = 10 seconds

To calculate:

• Retardation

Solution:

• Retardation = -0.416 m/s sq.

Using concepts:

• Formula to convert kmph-mps.

• To find the retardation we can use either acceleration formula or first equation of motion.

Using formulas:

• Formula to convert kmph-mps:

• ${\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}$

• Acceleration formula:

• ${\sf{a \: = \dfrac{v-u}{t}}}$

• First equation of motion:

• ${\sf{v \: = u \: + at}}$

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

Required solution:

~ Firstly let us convert initial velocity into metre per second!

$:\implies \sf 60 \times \dfrac{5}{18} \\ \\ :\implies \sf \dfrac{300}{18} \\ \\ :\implies \sf 16.66 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

~ Now let us convert final velocity into metre per second!

$:\implies \sf 45 \times \dfrac{5}{18} \\ \\ :\implies \sf \dfrac{225}{18} \\ \\ :\implies \sf 12.5 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

• Initial velocity = 16.66 mps
• Final velocity = 12.5 mps

~ Now calculating retardation!

By using first equation of motion.

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 12.5 \: = 16.66 + a(10) \\ \\ :\implies \sf 12.5 - 16.66 = 10a \\ \\ :\implies \sf -4.16 = 10a \\ \\ :\implies \sf \dfrac{-4.16}{10} \: = a \\ \\ :\implies \sf -0.416 \: = a \\ \\ :\implies \sf a \: = -0.416 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -0.416 \: ms^{-2}$

By using acceleration formula.

$:\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{12.5-16.66}{10} \\ \\ :\implies \sf \dfrac{-4.16}{10} \: = a \\ \\ :\implies \sf -0.416 \: = a \\ \\ :\implies \sf a \: = -0.416 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -0.416 \: ms^{-2}$

Answer 2

Question:

• A car moving with uniform velocity 60 km/hr retards in 10 second and reduces to 45 km/hr .find the retardation. ​

Answer:

• The retardation of the car is - 0.41 m/s²

Explanation:

Given that:

• A car moving with uniform velocity 60 km/hr
• It retards in 10 second and reduces to 45 km/hr

To Find:

• The value of the retardation of the car

Formula Used:

$\bigstar \; \; {\underline{\boxed{ \bold { Retardation = \dfrac{\triangle V }{\triangle T} }}}}$  

Required Solution:

• using formula to find retardation

★ We know that ,

${\pink{\bigstar \; {\boxed{\bf{ Retardation = \dfrac{v_{(Final)} - v_{(Initial)}}{t_{(time)}} }}}}}$  

★ Where ,

• Velocity is considered as rate of change of position
• Retardation is considered as the negative rate of change in velocity
• Time period is the progression of events from the past to the present into the future

★ Here we know that ,

• Final Velocity is 45 kmph
• Initial Velocity is 60 kmph
• Time period is 10 seconds

★ Converting values to S.I system

${\purple {\bigstar \; \; {\underline {\bf { Initial \; Velocity : }}}}}$

$\\ \longrightarrow \rm Velocity_{(Initial )} = 60 km/hr \\ \\ \\ \longrightarrow \rm Velocity_{(Initial )} = 60 * \dfrac{5}{18} \\ \\ \\ \longrightarrow \rm Velocity_{(Initial )} = 10 * \dfrac{5}{3} \\ \\ \\ \longrightarrow \rm {\red{\underline{\underline{Velocity_{(Initial )} = 16.6 \;m/s }}}}$

★ Henceforth ,

• The initial velocity of the car is 16.66 m/s

${\purple {\bigstar \; \; {\underline {\bf { Final \; Velocity : }}}}}$

$\\ \longrightarrow \rm Velocity_{(Final )} = 45 \; km/hr \\ \\ \\ \longrightarrow \rm Velocity_{(Final )} = 45 * \dfrac{5}{18} \\ \\ \\ \longrightarrow \rm Velocity_{(Final)} = 15 * \dfrac{5}{6} \\ \\ \\ \longrightarrow \rm {\red{\underline{\underline{Velocity_{(Final )} = 12.5 \;m/s }}}}$

★ Henceforth ,

• The final velocity of of the car is 12.5 m/s

According to the question:

• As we know thee initial velocity , the final velocity and the time taken by the car to reach the retardation now let's apply suitable formulae and find out the retardation of the car with the help of the given terms

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${\purple {\bigstar \; \; {\underline {\bf { Finding \; retardation: }}}}}$  

$\\ \longrightarrow \rm Retardation _{(of \; the \; car)} = \dfrac{\triangle v }{\triangle t} \\ \\ \\ \longrightarrow \rm Retardation _{(of \; the \; car)} = \dfrac{12.5 m/s - 16.6 m/s}{10 \;s} \\ \\ \\ \longrightarrow \rm Retardation _{(of \; the \; car)} = \dfrac{-4.1 m/s}{10 \; s} \\ \\ \\ \longrightarrow \rm {\red{\underline{\underline{Retardation _{(of \; the \; car)} = - 0.41 m/s^2 }}}}$

★  Henceforth ,

• The retardation of the car is - 0.41 m/s²

Therefore:

• The retardation or deceleration of the car is - 0.41 m/s²

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