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A car accelerates from $10 \text{ m/s}$ to $ 60 \text{ m/s}$ at an acceleration of $10 \text{ m/s$^2$}$. Calculate the time taken for this acceleration to occur and the distance the car has traveled in time.

I know that the acceleration is $10 \text{ m/s$^2$}$ and the initial velocity is $10 \text{ m/s}$ and the final velocity is $60 \text{ m/s}$.

However, I am unsure whether to use the equation $v = u + at$ or the equation $s = ut + \frac {1}{2}at^2$.

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You were on the right track. You know that $u=10$, $v=60$, $a=10$

In the equation v=u+at the only thing you're missing is time so that formula can be used to get the time taken.

The other formula will be useful once you know the time taken.

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  • $\begingroup$ So the final answers I got were time taken is 5[s] and the distance is 175[m^2], however it is asking for the distance the car has travelled in time so would I have to convert the 175[m^2] to time ? And if so how would I do this ? Thanks. $\endgroup$ – Dan Oct 19 '16 at 14:20
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v = u + at is the equation you will want to use.

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1) to calculate time t use first law of motion, $v=u+at$, $$60=10+10t$$ $$t=\frac{50}{10}=5 sec$$ 2) to calculate distance s traveled in time $t=5sec$ use 2nd law of motion, $s=ut+\frac12 at^2$, $$s=10\cdot 5+\frac12 \cdot 10\cdot 5^2=175 m$$

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  • $\begingroup$ For the second equation should the answer not be 175[m^2] rather than 140[m]? $\endgroup$ – Dan Oct 19 '16 at 14:31
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    $\begingroup$ i am sorry, you are correct, it is 175 m, again sorry for this silly mistake $\endgroup$ – jeanne clement Oct 19 '16 at 14:35
  • $\begingroup$ It's fine haha but I just wanted to ask can you leave the answer for distance in units of [m] as the question says the distance the car has travelled in time. Would you have to convert this into time or could you just leave the answer as it is ? Thanks. $\endgroup$ – Dan Oct 19 '16 at 14:44
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    $\begingroup$ distance traveled in time is the distance which car traveled in 5 sec i.e. time required to accelerate from 10 m/s to 60 m/s. read carefully the accepted answer. he is also saying the same thing that's first you calculate time then distance traveled in that time using your second equation $\endgroup$ – jeanne clement Oct 19 '16 at 14:52

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