The fast and the curious

If the initial velocity is $v_0$ , total distance traveled is $d$ , the taken time is $t$ and the acceleration is $a$ , then...

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$\Large{d=v_0 t+\frac {1}{2} at^2}$

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We have the values...

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$\large{v_0 = 0}$

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$\large{d=\frac {1}{4} \textrm{miles}}$

$~~~~\large{= (\frac {1}{4} \times 1.609 \times 1000) \textrm{meters}}$

$~~~~\large{= 402.25 ~ \textrm{meters}}$

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$\large{t=10 ~ \textrm{seconds}}$

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Plugging in the values, we get...

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$\large{402.25 ~ \textrm{meters}=0+\frac{1}{2}\times a \times (10~\textrm{second})^2}$

$\large{a=}$ $\Large{\frac{402.25 ~ \textrm{meters}}{50 ~ \textrm{second}^2}}$

$\Large{\therefore a=8.045 ~ \textrm{meter/second}^2}$

From $s=ut + \frac{1}{2}at^2$ ,

where $s=\frac{1}{4}mile=\frac{1}{4}(1.609)km, u=0, a=?, t=10$

substitute inside and we get $8.045m/s^2$

Hello all,

Given that the s = 0.25 mile = 0.25 (1.609)(1000) = 402.25 m, u = 0 m/s,

By applying the formula,

s = ut + 0.5at^2

402.25 = 0(10) + (0.5)(a)(10)^2

50a = 402.25

a = 8.045 m/s/s...

thanks...

We know that, s=ut+(0.5)a*t^2 (here we assume u as the initial velocity , a as the accelaration and t is the time elapsed and s is the distance it has gone )

Now, the car was started from rest so u=0 and s=1/4mile =0.40225KM=402.25m and t=10s now , using the information on the formula we will get that ,

402.25=(0.5)a*t^2 =>a=8.045m/s^2

1/4 mile of a mile= 1/4*1609 meters =402.25 meters

u=0 a=? (not known) t=10 seconds s=402.25 meters

s=ut+1/2*at²

402.25= 0+1/2*a100

4.02=1/2*a

a=2*4.02

a=8.04

• S = 1/4 mile = 1609/4 meters = 402.5 meters
• S = $Vo^{2}$ $\times t$ + 1/2 a $t^{2}$ since the Vo = 0, so
• S = 0 + 1/2 a $10^{2}$
• 402.5 = $50 \times a$
• a = 402.5/50
• a = 8.045