Question

brakres applied to train moving with velocity of 20 m/s if the train come to halt after 10 seconds calculate the acceleration of the train the distance covered by the train after the break are applied​

Answer 1

Answer:

Explanation:

u=20m/s,v=0,t=5s

v=u+at

0=20+a×5

a=−4m/s  

2

 

Therefore, Retardation=4m/s2

PLEASE THANK AND MARK BRAINLIST

 

 

Answer 2

Answer:

Provided that:

• Initial velocity = 20 mps
• Final velocity = 0 mps
• Time = 10 seconds

To calculate:

• The acceleration
• The distance covered

Solution:

• The acceleration = -2 mps²
• The distance covered = 100 m

Using concept:

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

• To calculate the distance we can use either second equation of motion or third equation of motion.

Using formulas:

• Acceleration is given by

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

• First equation of motion given by

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

• Second equation of motion given by

• ${\small{\underline{\boxed{\pmb{\sf{s \: = ut \: + \dfrac{1}{2} \: at^2}}}}}}$

• Third equation of motion is given by

• ${\small{\underline{\boxed{\pmb{\sf{v^2 \: - u^2 \: = 2as}}}}}}$

Required solution:

~ Firstly let us find out the acceleration!

By using acceleration formula

$:\implies \sf a \: = \dfrac{v-u}{t} \\ \\ :\implies \sf a \: = \dfrac{0-20}{10} \\ \\ :\implies \sf a \: = \dfrac{-20}{10} \\ \\ :\implies \sf a \: = -2 \: mps^{-2} \\ \\ :\implies \sf Acceleration \: = -2 \: mps^{-2} \\ \\ :\implies \sf Retardation \: = -2 \: mps^{-2}$

By using first equation of motion

$:\implies \sf v \: = u \: + at \\ \\ :\implies \sf 0 = 20 + a(10) \\ \\ :\implies \sf 0 = 20 + 10a \\ \\ :\implies \sf 0 - 20 = 10a \\ \\ :\implies \sf -20 = 10a \\ \\ :\implies \sf a \: = \dfrac{-20}{10} \\ \\ :\implies \sf a \: = -2 \: mps^{-2} \\ \\ :\implies \sf Acceleration \: = -2 \: mps^{-2} \\ \\ :\implies \sf Retardation \: = -2 \: mps^{-2}$

~ Now let's find out the distance travelled!

By using third equation of motion

$:\implies \sf v^2 \: - u^2 \: = 2as \\ \\ :\implies \sf (0)^{2} - (20)^{2} = 2(-2)(s) \\ \\ :\implies \sf 0 - 400 = -4s \\ \\ :\implies \sf -400 = -4s \\ \\ :\implies \sf 400 = 4s \\ \\ :\implies \sf \dfrac{400}{4} \: = s \\ \\ :\implies \sf 100 \: = s \\ \\ :\implies \sf Distance \: = 100 \: m$

By using second equation of motion

$:\implies \sf s \: = ut \: + \dfrac{1}{2} \: at^2 \\ \\ :\implies \sf s \: = 20(10) + \dfrac{1}{2} \times -2(10)^{2} \\ \\ :\implies \sf s \: = 200 + \dfrac{1}{2} \times -2(100) \\ \\ :\implies \sf s \: = 200 + \dfrac{1}{2} \times -200 \\ \\ :\implies \sf s \: = 200 + 1 \times -100 \\ \\ :\implies \sf s \: = 200 - 100 \\ \\ :\implies \sf 100 \: = s \\ \\ :\implies \sf Distance \: = 100 \: m$