Question

3. A car moving along a straight highway with a
speed of 72 kmh -- is brought to a stop within a
distance of 100 m. What is the retardation of the car
and how long does it take for the car to stop ?



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Answer 1

Answer :

Retardation of car is 2m/s² and time taken for the car to stop is 10sec

Explanatìon :

Given :

Initial velocity of a car ,u = 72km /hr = 20 m/s

Final velocity of car , v = 0 m/s

Distance covered , S= 100 m

To Find :

• Retardation
• Time taken for the car to stop

${\purple{\boxed{\large{\bold{Formula's}}}}}$

Kinematic equations for uniformly accelerated motion .

$\bf\:v=u+at$

$\bf\:S=ut+\frac{1}{2}at{}^{2}$

$\bf\:v{}^{2}=u{}^{2}+2aS$

and $\bf\:s_{nth}=u+\frac{a}{2}(2n-1)$

${\underline{\sf{Answer}}}$

We have

• u = 20 m/s
• v= 0
• S = 100 m

By third Equation of Motion

$\rm\:v{}^{2}=u{}^{2}+2aS$

Put the given values

$\sf\implies\:0=(20)^2+2a\times100$

$\sf\implies\:-(20)^2=2a\times100$

$\sf\implies\:a=\dfrac{-20\times20}{2\times100}$

$\sf\implies\:a=\dfrac{-4\times100}{2\times100}$

$\sf\implies\:a=-2ms^{-2}$

Then ,

By using first Equation of motion

$\rm\:v=u+at$

Put the given values , then

$\sf\implies\:0=20-2t$

$\sf\implies\:2t=20$

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

Answer 2

Answer:

Answer :

Retardation of car is 2m/s² and time taken for the car to stop is 10sec

Explanatìon :

Given :

Initial velocity of a car ,u = 72km /hr = 20 m/s

Final velocity of car , v = 0 m/s

Distance covered , S= 100 m

To Find :

Retardation

Time taken for the car to stop

{\purple{\boxed{\large{\bold{Formula's}}}}}

Formula

′

s

Kinematic equations for uniformly accelerated motion .

\bf\:v=u+atv=u+at

\bf\:S=ut+\frac{1}{2}at{}^{2}S=ut+

2

1

at

2

\bf\:v{}^{2}=u{}^{2}+2aSv

2

=u

2

+2aS

and \bf\:s_{nth}=u+\frac{a}{2}(2n-1)s

nth

=u+

2

a

(2n−1)

{\underline{\sf{Answer}}}

Answer

We have

u = 20 m/s

v= 0

S = 100 m

By third Equation of Motion

\rm\:v{}^{2}=u{}^{2}+2aSv

2

=u

2

+2aS

Put the given values

\sf\implies\:0=(20)^2+2a\times100⟹0=(20)

2

+2a×100

\sf\implies\:-(20)^2=2a\times100⟹−(20)

2

=2a×100

\sf\implies\:a=\dfrac{-20\times20}{2\times100}⟹a=

2×100

−20×20

\sf\implies\:a=\dfrac{-4\times100}{2\times100}⟹a=

2×100

−4×100

\sf\implies\:a=-2ms^{-2}⟹a=−2ms

−2

Then ,

By using first Equation of motion

\rm\:v=u+atv=u+at

Put the given values , then

\sf\implies\:0=20-2t⟹0=20−2t

\sf\implies\:2t=20⟹2t=20

\sf\implies\:t=10sec⟹t=10sec