It will take the car 16.67 sec to catch up with the truck.
Given the data in the question.
Since the car was stationary, then;
Initial velocity of the car; [tex]u_c = 0m/s[/tex].
Final velocity of the car; [tex]v_c = 30m/s[/tex].
Time taken for the car to reach given speed; [tex]t = 10s[/tex].
At [tex]t = 0[/tex], speed of truck; [tex]V_{truck} = 25m/s[/tex]
First, we find the acceleration of the car, using the first equation of motion;
[tex]v = u + at[/tex]
We make "a" the subject of the formula,
[tex]a = \frac{v-u}{t}[/tex]
We substitute in our values and solve for "a"
[tex]a = \frac{30m/s - 0m/s}{10s}[/tex]
[tex]a = 3m/s^2[/tex]
Now, to determine how long will it take for the car to catch up with the truck.
Distance travelled by truck should be equal to distance travelled by car, so;
[tex]S_{truck[/tex] = [tex]S_{car[/tex]
We know that, speed; s = distance / time
So, distance of truck [tex]S_{truck[/tex] = speed of truck × time [tex]= V_{truck} * t[/tex]
Also, from the second equation of motion; distance "s" is;
[tex]s = ut + \frac{1}{2}at^2[/tex]
So distance of the car [tex]S_{car} = ut + \frac{1}{2}at^2[/tex]
We equate
[tex]S_{truck[/tex] = [tex]S_{car[/tex]
[tex]V_{truck} * t[/tex] [tex]= ut + \frac{1}{2}at^2[/tex]
we make "t" the subject of formula
we know that initial velocity of the car; u = 0
so, we divide both sides by "t"
[tex]V_{truck} = \frac{1}{2}at[/tex]
[tex]t = \frac{2V_{truck}}{a}[/tex]
We substitute in our values
[tex]t = \frac{2 * 25m/s}{3m/s^2}[/tex]
[tex]t = 16.67s[/tex]
Therefore, It will take the car 16.67 sec to catch up with the truck.
Learn more; https://brainly.com/question/17685166?referrer=searchResults
