Question

A car starts from rest and accelerates uniformly at 3.0 m/s2. A second car starts from rest 6.0 seconds later at the same point and accelerates uniformly at 5.0 m/s2. How long does it take the second car to overtake the first car?

Answer 1

5(3)(t+6)^2 = .5(5)(t)^2 
t=20.62 seconds 

20.62 seconds

Answer 2

Answer:

20.62 seconds

Step-by-step explanation:

Intially, the first car starts from rest, therefore its displacement through the first 6 seconds can be found as follows:

$d=\frac{at^{2} }{2} =\frac{3*6^{2} }{2}\\d= 54 m$

Then, we need to find out its velocity at this point, which would be the first car's initial velocity for the second moment, when the second car accelerates:

$v_{1} =at=3*6=18 m/s$

To find out the moment that the second car overtakes the first one we should equal both cars' displacement equations and solve for t, keep in mind that a 54 m head start should be added to the first car's displacement:

$v_{1}t + \frac{a_{1}t^{2}}{2}+54 =\frac{a_{2}t^{2}}{2}\\18t + \frac{3t^{2}}{2}+54 =\frac{5t^{2}}{2}\\t^{2} -18t -54 = 0$

Solving for t will yield a positive and a negative answer, since time can't be negative, only the positive answer should be considered, therefore, t = 20.62 s