Question

A car travelling at a velocity of 20m/s dedelerates to a velocity of 10 m/s in a time interval of 5s. calculate the deceleration​

Answer 1

Question: A car travelling at a velocity of 20m/s dedelerates to a velocity of 10 m/s in a time interval of 5s. Calculate the deceleration

Provided that:

• Final velocity = 10 mps
• Initial velocity = 20 mps
• Time = 5 seconds

To calculate:

• Deceleration

Solution:

• Deceleration = +2 mps sq.

Using concept:

• We can use either first equation of motion or acceleration formula to solve this question.

Using formula:

• Acceleration formula:

• ${\small{\underline{\boxed{\pmb{\sf{a \: = \dfrac{v-u}{t}}}}}}}$

• First equation of motion:

• ${\small{\underline{\boxed{\pmb{\sf{v \: = u \: + at}}}}}}$

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

Required solution:

By using acceleration formula...

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{10-20}{5} \\ \\ :\implies \sf a \: = \dfrac{-10}{5} \\ \\ :\implies \sf a \: = -2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -2 \: ms^{-2} \\ \\ :\implies \sf Deceleration \: = +2 \: ms^{-2}$

By using first equation of motion...

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 10 = 20 + a(5) \\ \\ :\implies \sf 10 - 20 = 5a \\ \\ :\implies \sf -10 = 5a \\ \\ :\implies \sf a \: = \dfrac{-10}{5} \\ \\ :\implies \sf a \: = -2 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -2 \: ms^{-2} \\ \\ :\implies \sf Deceleration \: = +2 \: ms^{-2}$

Answer 2

Question:

• A car travelling at a velocity of 20m/s dedelerates to a velocity of 10 m/s in a time interval of 5s. calculate the deceleration​

Answer:

• The deceleration or retardation of the car is  2m/s²

Explanation:

Given that :

• A car travelling at a velocity of 20m/s dedelerates to a velocity of 10 m/s in a time interval of 5

To Find:

• The value of deceleration of the car

Formula Used:

$\bigstar \; \; {\underline{\boxed{\bf{ Deceleration = \dfrac{\triangle V}{\triangle T} }}}}$

Required Solution:

• Using first equation of motion

${\pink{\bigstar \; \; {\boxed{ \bf{ Deceleration = \dfrac{u_{(Initial \; velocity )} - v_{(Final\; velocity)}}{T_{(Time)}} }}}}}$  

★ Where ,

• Velocity is considered as rate of change of position
• Deceleration 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 10m/s
• Initial Velocity is 20 m/s
• Time period is 5 seconds

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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${\pink{\bigstar \; \; {\underline{\bf{Finding \; Deceleration : }}}}}$

$\\ \longrightarrow \sf Deceleration _{(of \; the \; car)} = \dfrac{u_{(Initial \; velocity )}- v _{(Final \; velocity)}}{t_{(time \; interval)}} \\ \\ \\ \ \longrightarrow \sf Deceleration _{(of \; the \; car)} = \dfrac{20 m/s - 10m/s}{5 \; s} \\ \\ \\ \ \ \longrightarrow \sf Deceleration _{(of \; the \; car)} = \dfrac{ 10m/s}{5 \; s} \\ \\ \\ \ \longrightarrow \sf {\red{\underline{\underline{Deceleration _{(of \; the \; car)} =2 m/s }}}}$  

★ Henceforth ,

• The deceleration of the car is =  2m/s

Therefore :

• The deceleration or retardation of the car is 2m/s²

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