Question

A bus starts from rest and moves down a hill with constant acceleration. The bus travels a distance of 400 m in 10 seconds. If mass of the bus is 5 metric tonne, then what will be its acceleration and force acting on it?​

Answer 1

$\sf Given \begin{cases} & \sf{Initial\; Velocity, u = \bf{0\;m/s}} \\ & \sf{Distance, d = \bf{400\;m} } \\ & \sf{Time, t = \bf{10\;s}} \\ & \sf{Mass, m = 5\;metric\;tonne = \bf{5000\; kg}}\end{cases}$

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☯ By using 2nd equation of motion,

$\star\;{\boxed{\sf{\purple{s = ut + \dfrac{1}{2} at^2}}}}$

$:\implies\sf 400 = 0 \times 10 + \dfrac{1}{2} a(10)^2$

$:\implies\sf 400 = 0 + \dfrac{1}{2} \times 100a$

$:\implies\sf 400 = \dfrac{1}{2} \times 100a$

$:\implies\sf 400 = 50a$

$:\implies\sf a = \cancel{ \dfrac{400}{50}}$

$:\implies{\boxed{\pink{\sf{a ={\frak{ 8\;{\sf{m/s^2}}}}}}}}\;\bigstar$

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☯ Now, As we know,

$\star\;{\boxed{\sf{\purple{F = ma}}}}$

$:\implies\sf F = 5000 \times 8$

$:\implies{\boxed{\pink{\sf{F ={\frak{ 40,000\;{\sf{N}}}}}}}}\;\bigstar$

$\therefore\;{\sf{Hence,\;Force\;acting\;on\;body\;is\;40,000\;N.}}$

Answer 2

Given :-

initial velocity, u = 0m/s

distance, s = 400m

time taken, t = 10 seconds

mass, metric tonne (m) = 5 metric tonne = 5000kg

Formula used :-

s = ut + 1/2at²

Now putting the values in :-

$\sf 400 = (0 \times 10) + ( \frac{1}{2} \times a {10}^{2} )$

$\sf 400 = 0 + \frac{1}{2} \times a \times 100$

$\sf 400 = 1 \times a \times 50$

$\sf 400 = 50a$

$\sf \frac{400}{50} = a$

$\sf 8 = a$

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Therefore,

the acceleration acquired here is of 8m/s²

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Now, force applied :-

Formula used :-

F = m*a

putting values in it :-

F = m(5000) * a(8)

F = 5000*8

F = 40,000

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Therefore, the force applied here is 40,000N