Respuesta :

Iqta

Hi!


Answer: 3.45 seconds


First you note down the known values:

s= 60m

v=0

t= ?

a= -10m/s^2


The equation of motion linking these values in a manner as to obtain the unknown value of t is:

s = vt - 1/2at^2

rearranging the equation for time,

t^2 = ( vt - s ) x 2 / a

t^2 = (0t - 60) x 2 / -10m/s^2

t^2 = 12

t = 3.45 s



alinakincsem

Answer:

3.46 seconds

Explanation:

We are given that a car stops in 60 m and has an acceleration of -10m/s[tex]^2[/tex] and we are to find the time it took to stop.

To find this time, we will use this equation of motion:

[tex] s = vt - \frac {1} {2} a t^2 [/tex]

Rearranging the formula by making t the subject and then substituting the given values in the above formula to find t:

[tex]t^2=(vt-s)*\frac{2}{a}[/tex]

[tex]t^2 = (0)t - 60 * \frac{2}{-10}[/tex]

[tex]\sqrt{t}^2=\sqrt{12}[/tex]

[tex]t=3.46[/tex]

Therefore, it took 3.46 seconds for the car to stop.

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