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STALIN [3.7K]
3 years ago
15

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?
Physics
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

#LearnwithBrainly

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