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A car accelerates from rest for $15$ s with a uniform acceleration of $1.5$ m/s^2 and immediately decelerates with a uniform deceleration of $5$ m/s^2.

How long does deceleration take?

I used $v=u+at$ to try find to the time taken but it's turning out right.

Lonidard
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sushi181
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3 Answers3

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I don't know the equations but I do know that the speed would be an arithmetic sequence and you can take the average speed for each of the 15 seconds and multiply it by 15 to get the distance. Then you do the same with the 5m/s but instead u set it up as an equation.

Angela
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First of all, this question is more suitable for physics forums. Try to consider it for next time ;)

In order to answer your question, you have two equations of motion for the car. First, between $t=0$ and $t=15$ the equation for the velocity is $v=v_0+a*t$. As you said nothing about the initial velocity of the car, I will suppose it is 0. The final velocity is then $v(t=15)=1.5*15=22.5m/s$.

Then the car starts to decelarate, and the equation we have is $v_2=v(t=15)+at=22.5m/s-5*t$. As we want the time when the velocity is 0, we impose $v_2=0$, finding that the deceleration time is equal to $t_{dec}=22.5/5s=4.5s$

user3209698
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Using $v=u+at$ for acceleration, we get $v=0+(1.5)(15)=22.5m/s$.

Again, for the deceleration half, again use $v=u+at$, but $v=0$( it ultimately stops) and $a=(-5)m/s^2$(-ve because it is retardation), and $u=22.5$(i.e. the deceleration starts from the velocity at the end of acceleration period, which is 22.5 as shown above)

So $0=22.5-5t$; giving $t=4.5$ seconds.

GRrocks
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