Question

A car accelerates uniformly from 18km h a car accelerates uniformly from 18 km per hour to 36 km per hour in 5 seconds calculate the acceleration and the distance covered by the car in that time

Answer 1

u(initial velocity) =18x 5/18m/s
=5m/s
v(final velocity) =36x 5/18
=10m/sacceleration (a) =10-5/5
1m/s
we know
s=ut + 1/2att
s=25+1/2x25
s=25+12.5
s=37.5m

Answer 2

Given :-

• Initial velocity, u = 18 km/h
• Final velocity, v = 36 km/h
• Time taken, t = 5 seconds

To Find :-

• Acceleration, a
• Distance covered,s

Formula to be used:-

• 1st equation of motion, v = u + at
• 3rd equation of motion, v² - u² = 2as

Solution :-

We are given :-

• Initial velocity of bus , u= 18 km/h.
• Final velocity of bus , v =36 km/h

$\qquad$☀So, firstly we have to change the km/h into m/s.For this we need to multiply by 5/18.

$\sf \purple{:\implies Initial\: velocity\: (u) = 18 \: km/h}$

$\sf \: \: \: \: \: \: \:\: \: :\implies Initial \: velocity = 18 \times \dfrac {5}{18}$

$\sf \: \: \: \: \: \: \:\: \: :\implies Initial \: velocity = \dfrac {90}{18}$

$\sf\purple{ \: \: \: \: \: \: \:\: \: :\implies Initial \: velocity = 5\: m/s}$

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$\sf \green{:\implies Final\: velocity\: (u) = 36 \: km/h}$

$\sf \: \: \: \: \: \: \:\: \: :\implies Final \: velocity = 36 \times \dfrac {5}{18}$

$\sf \: \: \: \: \: \: \:\: \: :\implies Final \: velocity = \dfrac {180}{18}$

$\sf\green{ \: \: \: \: \: \: \:\: \: :\implies Final \: velocity = 10\: m/s}$

$\qquad\small\underline{\pmb{\sf Substituting \: the\: given\: values \::-}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{ v = u + at}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ 10 = 5 + a × 5 }}$

$\qquad\leadsto\quad \pmb {\mathfrak{ 10 - 5 = 5a}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ 5 = 5a}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ \cancel{\dfrac{5}{5}}= a}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{ a = 1 m/s².}}}$

Hence, the acceleration of the car is 1 m/s².

$\qquad\small\underline{\pmb{\sf Substituting \: the\: given\: values \::-}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{ v² - u² = 2as}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ (10)² - (5)² = 2 × 1× s}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ 100 -25 = 2s}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ 75 = 2s}}$

$\qquad\leadsto\quad \pmb {\mathfrak{ \cancel{\dfrac{ 75}{2 }}= s}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{ s = 37.5 m }}}$

Hence, the distance covered by car is 37.5 m.

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$\qquad$☀️Equations of motion :-

$\qquad$ ⑴ v = u + at

$\qquad$ ⑵ s = ut + ½at²

$\qquad$ ⑶ v² - u² = 2as

• v = final velocity
• u = initial velocity
• s = displacement
• t = time taken
• a = acceleration

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