Question

a car increase its speed from 20 kmph to 50kmph in 10 seconds what is the exeleration covered by the car​

Answer 1

$\maltese\:\underline{\textsf{\textbf{AnsWer :}}}\:\maltese$

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✒ Initial velocity (u) of the car is 20 kmph.

✒ Final velocity (v) of the car is 50 kmph.

✒ Time Interval (t) is 10 seconds.

✒ We need to find the Acceleration (a) of the car.

✒ First of all we need to convert the units of initial velocity and final velocity from kmph to mps.

✒ After making the units same put the all known values in first kinematical equation of motion.

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✪ CALCULATION ✪

$\longrightarrow\:\sf Initial \: velocity (u) = 20 \: km/h$

$\longrightarrow\:\sf Initial \: velocity (u) = 20 \times \dfrac{5}{18}$

$\longrightarrow\:\sf Initial \: velocity (u) = \dfrac{100}{18}$

$\longrightarrow\: \underline{ \underline{\sf Initial \: velocity (u) = 5.6 \: {ms}^{ - 1}}}$

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$\longrightarrow\:\sf Final \: velocity (v) = 50 \: km/h$

$\longrightarrow\:\sf Final \: velocity (v) = 50 \times \dfrac{5}{18}$

$\longrightarrow\:\sf Final \: velocity (v) = \dfrac{250}{18}$

$\longrightarrow \: \underline{ \underline{\sf Final \: velocity (v) = 13.9 \: {ms}^{ - 1} }}$

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☉ Concept :- Change in velocity of the particle per unit time called as Acceleration.

Apply the first kinematical equation of motion and plug in the known quantities in that equation.

$\longrightarrow\:\sf v = u + at$

$\longrightarrow\:\sf 13.9 = 5.6 + a \times 10$

$\longrightarrow\:\sf 13.9 - 5.6 = a \times 10$

$\longrightarrow\:\sf 8.3= a \times 10$

$\longrightarrow\:\sf a = \frac{8.3}{10}$

$\longrightarrow\: \boxed{\sf a = 0.83 \: {ms}^{ - 2} } \: \dag$

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Answer 2

$\huge\sf \orange\bigstar \: \underline {Question} :-$

A car increase its speed from 20 km/h to 50 km/h in 10 seconds what is the acceleration covered by the car ?

Given :-

• Car increase its speed from 20 km/h to 50km/h in 10 seconds.

To find :-

• Find is the acceleration covered by the car.

$\huge\sf \blue\bigstar \: \underline { Solution} : -$

$\sf \: u = 20 km/h \: \implies 20 \times \frac{5}{18} = 5.56 \: m/s$

$\sf \: v = 50 km/h \:\implies50 \times \frac{5}{18} = 13.89 \: m/s$

$\sf \: t = 10 \: sec$

$\sf \underline{ \red{Formula \: used}} : -$

$\sf \purple\implies Acceleration = \sf\frac{( v - u ) }{t}$

$\sf \purple \implies \frac{( 13.89 - 5.56 )}{10}$

$\: \: \: \: \: \: \boxed{\boxed{\sf0.833 \: {m/s}^{2}}}$

Some other information :-

• Acceleration is the rate of change of velocity.

• Usually, acceleration means the speed is changing, but not always.