Answer :
First we have to calculate the acceleration of the car,
[tex]a =\frac{u-v}{t}[/tex]
Here, u is initial velocity of the car and its value is given 12.2 m/s and v is final velocity of the car and it comes to stop, so its value zero.
[tex]a=\frac{0-12.2 m/s}{t} =\frac{-12.2 \ m/s}{t}[/tex].
As during the braking the acceleration is constant, from the kinematic equation,
[tex]s=ut + \frac{1}{2} a t^2[/tex]
Here, s is the distance traveled by the car during braking and its value is given 36. 5 m.
Substituting all the values in kinematic equation, we get
[tex]36.5 m =(12.2 m/s) t + \frac{1}{2} (- \frac{12.2 m/s}{t}) t^2 \\\\ 36.5 \ m = (12.2 \ m/s) t -(6.1 \ m/s) t \\\\\ t = \frac{36.5 m}{6.1 m/s} = 5.98 s[/tex]
Therefore, car will stop after 5.98 s.
Final answer:
The car took 6 seconds to come to a stop.
Explanation:
To calculate the time it took for the car to stop, we can use the equation of motion:
vf² = vi² + 2ad
where:
- vf is the final velocity (which is 0 m/s when the car comes to a stop)
- vi is the initial velocity (12.2 m/s)
- a is the acceleration (which is negative, since the car is decelerating)
- d is the distance covered (36.5 m)
Plugging in the given values, we have:
0² = 12.2² + 2a(36.5)
Simplifying the equation:
0 = 148.84 + 73a
Solving for a:
a = -2.04 m/s²
Now, we can use the formula:
a = (vf - vi) / t
to find the time t it took for the car to stop.
Plugging in the values:
-2.04 = (0 - 12.2) / t
Simplifying the equation:
t = 6 seconds
Learn more about Calculating time of deceleration here:
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