Answer:
d = 4217 m
Explanation:
Case 1.
Initial velocity, u = 0
Acceleration, a = 2.2 m/s²
Let d₁ is the distance in case 1. Using second equation of motion as follows :
[tex]d_1=ut+\dfrac{1}{2}a_1t_1^2\\\\\text{Here, u =0}\\\\d_1=\dfrac{1}{2}\times 2.2\times 10^2\\\\d_1=110\ m[/tex]
Let d₂ in the distance in case 2. Using second equation of motion as follows :
[tex]d_2=ut+\dfrac{1}{2}a_2t_2^2\\\\\text{Here, u =0}\\\\d_2=\dfrac{1}{2}\times 1.5\times 74^2\\\\d_2=4107\ m[/tex]
Total distance,
D = d₁ + d₂
D = 110 + 4107
D = 4217 m
Hence, the total distance covered is 4217 m.
