The Original question is as follows:
Two cars $A$ and $B$ simultaneously start a race. Velocity $v$ of the car $A$ varies with time $t$ according to graph shown. It acquires a velocity of $50 m/s$ few seconds before $t=100s$ and thereafter moves with this speed. Car $B$ runs together with car $A$ till both acquire a velocity of $20m/s$, after this car $B$ moves with zero acceleration for one second and then follows $v-t$ profile identical to car $A$ with a delay of $1$ second. In this way, car $B$ acquires a velocity of $50m/s$ one second after car$ A$ acquires it. How much more distance($\Delta s$) does car $A$ cover in first $100s$ as compared to car $B$ ?
My efforts:
Let both the cars move for $t_0$ seconds till they both acquire velocity of $20 m/s$. With the help of question, I draw the curve for car $B$ alongside the car $A$ (shown below). Hence, the answer to the question is the shaded area as shown below.
Let $S_A$ be the area (distance) travelled by car $A$ between the time interval $[t_0, 100]$ and $S_B$ be the area (distance) travelled by car $B$ between the time interval $[t_0, 100]$ . Then required area is $S_A – S_B$.
Now,
$$S_A = \int_{t_0}^{100}v_A(t)dt$$
and
$$S_B= 20 + \int_{t_0+1}^{100}v_B(t)dt$$
The $20$ in $S_B$ accounts for the rectangular area where car $B$ travels with zero acceleration.
Also, according to question, $v_B(t) = v_A(t-1)$ because car $B$ is delayed by $1$ second.
Hence, the answer($\Delta s$)should be :
$$\Delta s= \int_{t_0}^{100}v_A(t)dt – \int_{t_0+1}^{100}v_A(t-1)dt-20$$
I didn't included the distance travelled between $t=0$ to $t=t_0$ because they are same for both cars and will cancel out.
I'm stucked here as the integrals don't seem to be solved because nothing is known about the exact expression for $v_A(t)$. How can I proceed further and if there's something wrong I did ?
(P.S.: According to answer key, $\Delta s=30m$)
Best Answer
Note this key bit of information: "It acquires a velocity of $50$ m/s few seconds before $t=100$ s".
Let's rearrange B's speed profile. Take that second where B drives a constant $20$ m/s, and move it to the end:
This doesn't change the area under the curve for B. It just rearranges it. Thus the distance travelled is still the same. But now the speed profile for B is identical to that of A for the first 99 seconds. For that hundredth second, A has already reached its max speed of $50$ m/s, while B has suddenly dropped to $20$ m/s, so there is a $1$ second period with a $30$ m/s difference in their speed. Thus the distance travelled is $30$ m less for B.
You can also discover this from your integral equation: $$\Delta s= \int_{t_0}^{100}v_A(t)dt - \int_{t_0+1}^{100}v_A(t-1)dt-20$$ In the second integral, we make the substitution $t' = t-1$ and we get $\int_{t_0}^{99}v_A(t')dt'$ and so $$\begin{align}\Delta s &= \int_{t_0}^{100}v_A(t)dt - \int_{t_0}^{99}v_A(t')dt'-20 \\&= \int_{99}^{100}v_A(t)dt-20\\&=\int_{99}^{100}50dt-20\\&=30\end{align}$$