Question

a truck travelling at 54km /h is slow down to 36 km /h in 10 sec find the retardation of truck

pls only answer not explain
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Answer 1

Provided that:

• Initial velocity = 54 km/h
• Final velocity = 36 km/h
• Time taken = 10 seconds

To calculate:

• The retardation of truck

Solution:

• The retardation of truck = -0.5 m/s²

Using concepts:

• Acceleration formula
• Formula to convert kmph – mps

Using formula:

• Acceleration is given by,

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

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

Required solution:

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

$:\implies \sf 54 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{54} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 3 \times 5 \\ \\ :\implies \sf 15 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

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

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: ms^{-1} \\ \\ {\pmb{\sf{Henceforth, \: converted!}}}$

Therefore,

Initial velocity is 15 m/s

• Final velocity is 10 m/s

~ Now let's calculate retardation!

$:\implies \sf Acceleration \: = \dfrac{Change \: in \: velocity}{Change \: in \: time} \\ \\ :\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{10-15}{10} \\ \\ :\implies \sf a \: = \dfrac{-5}{10} \\ \\ :\implies \sf a \: = -0.5 \: ms^{-2} \\ \\ :\implies \sf Acceleration \: = -0.5 \: ms^{-2} \\ \\ :\implies \sf Retardation \: = -0.5 \: ms^{-2}$