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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

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

Learn more about uniformly accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

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