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3 years ago

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A car is moving 14.4 m/s. It decelerates at -1.55 m/s^2 for 4.26 s. It then accelerates at 1.83 m/s^2 until it reaches 11.8 m/s.

What was the total time?

1 answer:

Aleonysh [2.5K]3 years ago

7 0

The total time is 6.45 s

Explanation:

The motion of the car is a uniformly accelerated motion, so we can use suvat equations.

In the first part,

$u_1 = 14.4 m/s$ (initial velocity)

$a_1 = -1.55 m/s^2$ (acceleration)

$t_1=4.26 s$ (time of the first part)

So, we can find the velocity of the car after the first part, by using

$v_1 = u_1 +a_1 t_1 =14.4+(-1.55)(4.26)=7.8 m/s$

This is therefore the initial velocity of the second part:

$u_2 = v_1 = 7.8 m/s$

$a_2 = 1.83 m/s^2$ (acceleration in the second part)

$v_2 = 11.8 m/s$ (final velocity)

And therefore,

$v_2 = u_2 + a_2 t_2\\t_2=\frac{v_2-u_2}{a_2}=\frac{11.8-7.8}{1.83}=2.19 s$

So, the total time is

$t=t_1+t_2=4.26+2.19=6.45 s$

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The question is incomplete. Here is the complete question.

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v is velocity

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The equation will be:

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Car B started at the starting line. So, its equation is

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$D_{A}+v_{A}t=v_{B}t$

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Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

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$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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