Answer :
Answer:
The acceleration is [tex]1.106\ m/s^2[/tex] and the distance covered is 97.17 m.
Explanation:
Given that,
Initial speed of an automobile, u = 60 km/hr = 16.67 m/s
Final speed of an automobile, v = 80 km/hr = 22.2 m/s
Time, t = 5 s
We need to find the acceleration of the car and the distance traveled in this 5 sec interval. Let a is the acceleration. Using the definition of acceleration as :
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{22.2-16.67}{5}\\\\a=1.106\ m/s^2[/tex]
Let d is the distance covered. Using the third equation of motion to find it as follows :
[tex]v^2-u^2=2as\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{(22.2)^2-(16.67)^2}{2\times 1.106}\\\\s=97.17\ m[/tex]
So, the acceleration is [tex]1.106\ m/s^2[/tex] and the distance covered is 97.17 m.