Question

a bus travelling at a speed of 120 km breaks are appilied so as to produce a uniform acceleration of -4m/s2 a) find how for the bus will go before it in brought to rest​

Answer 1

Given:-

a bus travelling at a speed of 120 km/hr breaks are appilied so as to produce a uniform acceleration of -4m/s².

• Speed of train(u)=120km/hr=120x5/18=32m/s.
• Acceleration (a)= -4m/s²

To Find:-

how for the bus will go before it in brought to rest.

Means the Distance from where it bought to rest.

Solution:-

For Finding the distance, we will need to find time taken

So,

Formula Used will be :

$\mathcal\purple{\underbrace{\overbrace{\mid{\orange{\sf \: v = u + at}}\mid}}}$

it is given in the question that the train will bought to rest. So, The Final velocity(v)=0.

Now,

$0 = 32 + ( - 4) \times t$

$= > 0 = 32 - 4t$

$= > - 32 = - 4t$

$= > 4t = 32$

$= > t = \frac{32}{4}$

$= > t = 8s$

Time Taken=8s.

Time taken is 8s ,now we will find Distance by using Distance formula

i.e,

$\huge\underline{\overline{\mid{\bold{\red{\mathcal{\sf \: S = ut + \frac{1}{2} a {t}^{2} }}\mid}}}}$

where, S is the distance before the train is bought to rest.

Now,

$S = 32 \times 8 + \frac{1}{2} ( - 4) {8}^{2}$

$= > S = 256 + - (0.5 \times 4 \times 64)$

$= > S = (256 - 128)m$

$= > S = 128m$

∴the bus will go 128m before it in brought to rest.

Answer 2

Answer:

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