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erastova [34]
3 years ago
6

Two cars are traveling along a straight line in the same direction, the lead car at 24.7m/s and the other car at 29.9m/s. at the

moment the cars are 38.1m apart, the lead driver applies the brakes, causing her car to have an acceleration of -1.96m/s2. how long does it take for the lead car to stop? (in s)
Physics
1 answer:
sattari [20]3 years ago
7 0

Answer: 12.6 s

We would use the equation of motion:

v-u=at

where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

It is given that the lead car has an initial velocity, u=24.7\hspace{1mm} m/s. It decelerates with a=-1.96 \hspace{1mm} m/s^{2} to stop (v=0). we need to calculate the time taken by the lead car to stop.

\Rightarrow t=\frac{v-u}{a}\\ \Rightarrow t=\frac{0-24.7}{-1.96}=12.6\hspace{1mm} s

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<u> </u><u>»</u><u> </u><u>Image</u><u> </u><u>distance</u><u> </u><u>:</u>

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

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{ \tt{ \frac{1}{v} +  \frac{1}{10} =  \frac{1}{5}   }} \\  \\  { \tt{ \frac{1}{v}  =  \frac{1}{10} }} \\  \\ { \tt{v = 10}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \:  \: }}}}}

<u> </u><u>»</u><u> </u><u>Magnification</u><u> </u><u>:</u>

• Let's derive this formula from the lens formula:

{ \tt{ \frac{1}{v}  +  \frac{1}{u} =  \frac{1}{f}  }} \\

» Multiply throughout by fv

{ \tt{fv( \frac{1}{v} +  \frac{1}{u} ) = fv( \frac{1}{f}  )}} \\   \\ { \tt{ \frac{fv}{v}  +  \frac{fv}{u}  =  \frac{fv}{f} }} \\  \\  { \tt{f + f( \frac{v}{u} ) = v}}

• But we know that, v/u is M

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{ \tt{M =  \frac{10}{5} - 1 }} \\  \\ { \tt{M = 5 - 1}} \\  \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}

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