The traveling time of two cars

Question

Two cars $P , Q$ are moving in the same direction along the same straight horizontal road .Car P is moving with constant speed $25 m s^{-1}$ . At time $t = 0$ ,P overtakes Q which moving with constant speed $20 m s^{-1}$ . From $t= T$ seconds ,P decelerates uniformly , coming to rest at a point X which is $800 m$ from the point where P overtook Q . From $t= 25$ s , Q decelerates uniformly , coming to rest at the same point X at the same instant as P. Find the value of T ? My turn: For P $$ 800 = 25T + 25t + \frac{1}{2} a t^2 $$ For Q $$800 = 20 * 25 + 20 (T+t -25) + \frac{1}{2} a_1 (T+t -25)^2$$ But there is information about the acceleration of the two cars ?

Answer 1

Your equations for $P$ and $Q$ are not quite correct. Note the distance traveled when uniformly decelerating from traveling at speed $s$ m/s to $0$ in $t$ seconds is $\frac{ts}{2}$. To help visualize this, note if you had a graph of time on the $x$-axis and speed on the $y$-axis, the distance traveled would be the area under the graph. In this case, you have a straight line going from $y = s$ to $y = 0$ over a horizontal distance of $t$. Thus, you have right-angled triangle, so the area is, with base $b$ and height $h$, equal to $\frac{bh}{2}$, i.e., $\frac{ts}{2}$ here.

Thus, the $2$ equations are actually, letting $t_s$ be the time when both $P$ and $Q$ stop,

$$800 = 25T + (t_s - T)\frac{25}{2} = (t_s + T)\frac{25}{2} \tag{1}\label{eq1}$$ $$800 = 20 \times 25 + (t_s - 25)\frac{20}{2} = (t_s + 25)\frac{20}{2} \tag{2}\label{eq2}$$

Equating \eqref{eq1} and \eqref{eq2} gives

$$(t_s + T)\frac{25}{2} = (t_s + 25)\frac{20}{2} \implies t_s + T = \frac{4}{5}(t_s + 25) \implies T = \frac{-t_s}{5} + 20 \tag{3}\label{eq3}$$

From \eqref{eq2}, you have

$$800 = (t_s + 25)\frac{20}{2} \implies 80 = t_s + 25 \implies t_s = 55 \tag{4}\label{eq4}$$

Substituting this into \eqref{eq3} gives

$$T = \frac{-55}{5} + 20 = -11 + 20 = 9 \tag{5}\label{eq5}$$

Thus, $T = 9$ seconds is the answer.

Answer 2

It's a bit dangerous to use $t$ for the particular length of time one car is decelerating and also to use $t$ as a kind of generic "time" variable in several other places. But if you are consistent about using $t$ as the number of seconds in which the car decelerates from speed $25$ to speed zero, the acceleration of that car, $a$, satisfies the equation $$ at = 0 - 25 = -25.$$

If the other car decelerates from $20$ to zero in $T+t-25$ seconds, its acceleration $a_1$ must satisfy the equation $$ a_1 (T+t-25) = 0 - 20 = -20.$$

Solve these equations for $a$ and $a_1$ and you can use them to eliminate $a$ and $a_1$ from your other equations. Or simply identify the left sides of these equations in yours and replace them with the right sides. Either way, you will be able to simplify your equations considerably. (You should in fact end up with something equivalent to the main equations in the earlier answer, although with different notation.)