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 ???? after Stan does. Stan moves with a constant acceleration of 3.69 m/s while Kathy maintains an acceleration of 4.96 . (a) Find the time at which Kathy overtakes Stan. (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) 7.25 s

(b) 96.875 m

(c) 31 m/s

Explanation:

Given:

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

• Initial speed of Stan = $u_s$ = 0 m/s

• Constant acceleration of Kathy = $a_k = 4.96\ m/s^2$

• Constant acceleration of Stan = $a_s = 3.69\ m/s^2$

Assume:

• Time instant at which Kathy overtakes Stan = $t$

• Time interval for which Stan moves before Kathy overtakes him = $t_s = t$

• Time interval for which Kathy takes to overtake Stan =$t_k = t-1$
• Final speed of Stan when Kathy overtakes him = $v_s$

• Final speed of Kathy when she overtakes Stan = $v_k$

Part (a):

At the instant Kathy overtakes Stan, the distance traveled by Kathy must be equal to the distance traveled by Stan.

So, Distance traveled by Kathy = Distance traveled by Stan

Using the formula for constant acceleration motion, we have

$u_kt_k+\dfrac{1}{2}a_kt_k^2=u_st_s+\dfrac{1}{2}a_st_s^2\\\Rightarrow (0)(t-1)+\dfrac{1}{2}(4.96)(t-1)^2=(0)(t)+\dfrac{1}{2}(3.69)t^2\\\Rightarrow \dfrac{1}{2}(4.96)(t-1)^2=\dfrac{1}{2}(3.69)t^2\\\Rightarrow (4.96)(t-1)^2=(3.69)t^2\\\Rightarrow \dfrac{t^2}{(t-1)^2}=\dfrac{4.96}{3.69}\\\Rightarrow \dfrac{t^2}{(t-1)^2}=1.344$

Taking square root on both sides

$\dfrac{t}{(t-1)}=\sqrt{1.344}\\\Rightarrow \dfrac{t}{(t-1)}=1.16\\\Rightarrow t = (t-1)1.16\\\Rightarrow 0.16t = 1.16\\\Rightarrow t = 7.25$

Hence, Kathy overtakes Stan at t = 7.25 s.

Part (b):

Let us find the distance traveled by Kathy in this time interval.

$s_k = u_kt_k+\dfrac{1}{2}a_kt_k^2\\\Rightarrow s_k =(0)(t-1)+\dfrac{1}{2}(4.96)(t-1)^2\\\Rightarrow s_k =\dfrac{1}{2}(4.96)(7.25-1)^2\\\Rightarrow s_k = 96.875$

Hence, Kathy travels a distance of 96.875 m to overtake Stan.

Part (c):

Let us calculate the velocity of both Stan and Kathy at the instant when she overtakes him.

$v_k = u_k+a_kt_k\\\Rightarrow v_k = 0+(4.96)(t-1)\\\Rightarrow v_k = 0+(4.96)(7.25-1)\\\Rightarrow v_k =31$

Hence, the speed of Kathy is 31 m/s when she overtakes Stan.

$v_s = u_s+a_st_s\\\Rightarrow v_s = 0+(3.69)(t)\\\Rightarrow v_s = 0+(369)(7.25)\\\Rightarrow v_s =26.75$

Hence, the speed of Stan is 26.75 m/s when Kathy overtakes him.