Question

Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting line 1.00 s after Stan does. Stan moves with a constant acceleration of 3.9 m/s^2 while Kathy maintains an acceleration of 4.73 m/s^2. (a) Find the time at which Kathy overtakes (b) Find the distance she travels before she catches him (c) Find the speeds of both cars at the instant she overtakes him.

Answer 1

Answer:

(a) 10 s

(b) 236.5 m

(c) Kathy's speed = 47.3 m/s

    Stan's speed = 42.9 m/s

Explanation:

Given:

• $u_k$ = initial speed of Kathy = 0 m/s

• $u_s$ = initial speed of Stan = 0 m/s
• $a_k$ = acceleration of Kathy = $4.73\ m/s^2$

• $a_s$ = acceleration of Stan = $3.9\ m/s^2$

Assumptions:

• $v_k$ = final speed of Kathy when see catches Stan
• $v_s$ = final speed of Stan when Kathy catches him
• $s_k$ = distance traveled by Kathy to catch Stan
• $s_s$ = distance traveled by Stan when Kathy catches him
• $t_k$ = time taken by Kathy to catch Stan = $t$
• $t_s$ = time interval in which Kathy catches Stan = $t+1$

Part (a):

 Kathy will catch Stan only if the distances traveled by each of them are equal at the same instant.

$\therefore s_s=s_k\\\Rightarrow u_st_s+\dfrac{1}{2}a_st_s^2=u_kt_k+\dfrac{1}{2}a_kt_k^2\\ \Rightarrow (0)(t+1)+\dfrac{1}{2}(3.9)(t+1)^2=(0)(t)+\dfrac{1}{2}(4.73)t^2\\ \Rightarrow \dfrac{1}{2}(3.9)(t+1)^2=\dfrac{1}{2}(4.73)t^2\\\Rightarrow (3.9)(t+1)^2=(4.73)t^2\\\Rightarrow \dfrac{(t+1)^2}{t^2}=\dfrac{4.73}{3.9}\\\textrm{Taking square root in both sides}\\\dfrac{t+1}{t}= 1.1\\\Rightarrow t+1=1.1t\\\Rightarrow 0.1t = 1\\\Rightarrow t = 10$

Hence, Kathy catches Stan after 11 s from the Stan's starting times.

Part (b):

Distance traveled by Kathy to catch Stan will be distance the distance traveled by her in 10 s.

$s_s = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_s= (0)(t)+\dfrac{1}{2}(4.73)t^2\\\Rightarrow s_s= \dfrac{1}{2}(4.73)(10)^2\\\Rightarrow s_s= 236.5$

Hence, Kathy traveled a distance of 236.5 m to overtake Stan.

Part (c):

$v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.73)(t)\\\Rightarrow v_k = (4.73)(10)\\\Rightarrow v_k =47.3$

The speed of Kathy at the instant she catches Stan is 47.3 m/s.

$v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.9)(t+1)\\\Rightarrow v_s = (3.9)(10+1)\\\Rightarrow v_s =42.9$

The speed of Stan at the instant Kathy catches him is 42.9 m/s.