What formulas will solve problems like this? with both time and distance unknown?

Question

A car with an acceleration of 6m/s^2 exits the pit going 24m/s just as a car going 70m/s reaches the same location. If the car maintains it's acceleration, how long will it take to catch up with the other car?


Two men on horseback, 88 meters apart, accelerate toward each other. One going 0.2m/s^2 the other going 0.3m/s^2. Relative to the fastests man's starting point, when do they collide?

Answer 1

True, you're not given time nor distance; it's possible to obtain a solution, though. For, regardless of where you choose to place your reference point (the point you measure distance from), you know beforehand the distance traveled by each car is the same for both in at least two different times: 1- just when the first car A (that coming out of the pit) is passed by car B; 2- when car A catches up with car B. Setting t = 0 when car B overtakes car A as it runs by the pits, for car A,

s = vo t + ½ at² = 24 t + 3 t².

For car B,

s = v t = 70 t.

Since travesed distance is the same, let

s = 70 t = 24 t + 3 t².

Disregard s for the moment being; you are left with an equation in terms of t. Rearranging,

(70 − 24) t −3 t² = 0
46 t −3 t² = 0
(46 − 3 t) t = 0.

The solutions to the above equation are the instants at which both cars have traveled the same distance. Clearly, the expression above is zero when t = 0, the trivial solution (we knew that already). The other solution implies t = 46/3 = 15 1/3 s. You can now even determine the distance at which car A catches up with B: B will travel s = v t = 70 × 15 1/3 = 1 073 1/3 m ahead of the starting point of the first car before being overrun by A.

As for the second question, it doesn't make much sense to me. I think it should read "Relative to the fastest man's starting point, WHERE do they collide?". If they're asking WHEN do they collide, the choice of an spatial reference point is totally irrelevant. Anyhow, let it be the way they want it. Let x = 0 be the the fastest man's starting point. Assume point X is the collision point. Let t be collision time t. In this time, rider A (the fastest) will have covered X = ½ × 0.3 t² meters. The other rider travels the remaining distance, 88 − X, so

88 − X = ½ × 0.2 t².

Substituting X, as given in the first expression, into the second,

88 − ½ × 0.3 t² = ½ × 0.2 t².
(0.1 + 0.15) t² = 0.25 t² = 88
t = √(88 × 4) = 18.76 s.

This anwers WHEN do they collide. Clearly, collision point (point X) is located X = ½ × 0.3 t² = 52.8 m from fastest man's starting point.

Answer 2

Your teacher should probably have told you to make a table of kinematic variables for each body in the problem. I.e. write down x0, x, v0, v, a, and t for both cars in the first problem. Be sure to label them with "1"s for car #1 and "2" for car #2 so that you don't get them mixed up. Some variables you will know exactly, i.e. a1=6 m/s^2. Others you might not know exactly, like the final positions... but you do know the final positions are equal: x1=x2. For each table, write down the kinematic equation that involves only variables you know something about and the variable you want to know about (time, in this case). Solve the resulting system of equations.

Answer 3

1,) First car will overtake second car when his average velocity is 70m/s
V(avg) = (V(zero) + V(final))/2
2 V(avg) = V(zero) + V(final)
V(final) = 2V(avg) - V(zero)
V(final) = 2(70) - 24 = 116 m/s
So the first car's velocity went from 24m/s to 116m/s
His velocity changed by 116 - 24 = 92m/s
With an acceleration of 6m/s^2, that would have taken 92/6 = 15.33... seconds

2.)Combined speed is 0.2m/s + 0.3m/s = 0.5m/s
t = d/r = 88/0.5 = 176s
Relative to the fastest man's starting point, in 176 seconds he traveled:
D = rt = (0.3)176 = 52.8m