Complete Question
Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?
Answer:
Go-cart A is faster
Explanation:
From the question we are told that
The length of the track is [tex]l = 1.0 \ km = 1000 \ m[/tex]
The speed of A is [tex]v__{A}} = 20 \ m/s[/tex]
The uniform acceleration of B is [tex]a__{B}} = 0.333 \ m/s^2[/tex]
Generally the time taken by go-cart A is mathematically represented as
[tex]t__{A}} = \frac{l}{v__{A}}}[/tex]
=> [tex]t__{A}} = \frac{1000}{20}[/tex]
=> [tex]t__{A}} = 50 \ s[/tex]
Generally from kinematic equation we can evaluate the time taken by go-cart B as
[tex]l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2[/tex]
given that go-cart B starts from rest u = 0 m/s
So
[tex]1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2[/tex]
=> [tex]1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2[/tex]
=> [tex]t__{B}} = 77.5 \ seconds[/tex]
Comparing [tex]t__{A}} \ and \ t__{B}}[/tex] we see that [tex]t__{A}}[/tex] is smaller so go-cart A is faster
