Question

A car is moving with a speed of 90 km h-1
. The driver sees a road sign stating that there is
a block in the road in 1 km. He applies brake to produce a uniform acceleration of
-0.5 ms
-2
. Find how far before the block the car stops.

Answer 1

Given:

Initial velocity of car, u= 90 km h⁻¹

Final velocity of car, v= 0 km h⁻¹

(Since it stopped after applying brake)

Acceleration of car, a= -0.5 ms⁻²

Distance of Block from the point driver sees the sign, d= 1 km= 1000m    

To Find:

Distance from the block before which the car stops

Solution:

Let the distance covered by car after applying brake be 's' m

It is given that,

$\sf{u=90\:Km\:h^{-1}=90\times\dfrac{5}{18}\:ms^{-1}=25\:ms^{-1}}$

$\sf{v=0\:Km\:h^{-1}=0\times\dfrac{5}{18}\:ms^{-1}=0\:ms^{-1}}$

We know that,

• According to third equation of motion for constant acceleration

$\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}$  

where,

v is the final velocity of body

u is the initial velocity of body

a is acceleration of body

s is displacement of body

Now, on applying third equation of motion on given car, we get

$\longrightarrow\mathrm{v^{2}-u^{2}=2as}$

$\longrightarrow\mathrm{(0)^{2}-(25)^{2}=2(-0.5)s}$

$\longrightarrow\mathrm{-625=(-1)s}$

$\longrightarrow\mathrm{-625=-s}$

$\longrightarrow\mathrm{625=s}$

$\longrightarrow\mathrm{\green{s=625m}}$

Now,

Distance from the block before which the car stops

= d - s

= 1000 - 625

= 375 m= 0.375 Km

Hence, the car stops 375m or 0.375 Km before the block.

Answer 2

Answer:

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