Question

Two go-carts, A and B, race each other around a 1.0 track. Go-cart A travels at a constant speed of 20.0 /. Go-cart B accelerates uniformly from rest at a rate of 0.333 / ଶ . Which go-cart wins the race and by how much time?

Answer 1

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 $l = 1.0 \ km = 1000 \ m$

       The speed of  A is  $v__{A}} = 20 \ m/s$

       The uniform acceleration of  B is  $a__{B}} = 0.333 \ m/s^2$

  Generally the time taken by go-cart  A is mathematically represented as

              $t__{A}} = \frac{l}{v__{A}}}$

=>          $t__{A}} = \frac{1000}{20}$

=>           $t__{A}} = 50 \ s$

  Generally from kinematic equation we can evaluate the time taken by go-cart B as

             $l = ut__{B}} + \frac{1}{2} a__{B}} * t__{B}}^2$

given that go-cart B starts from rest  u =  0 m/s

So

            $1000 = 0 *t__{B}} + \frac{1}{2} * 0.333 * t__{B}}^2$

=>         $1000 = 0 *t__{B}} + \frac{1}{2} 0.333 * t__{B}}^2$            

=>         $t__{B}} = 77.5 \ seconds$  

 

Comparing  $t__{A}} \ and \ t__{B}}$  we see that $t__{A}}$ is smaller so go-cart A is  faster