Question

At a stop light, a truck traveling at 15 m/s passes a car as it starts from rest. The truck travels at constant velocity and the car accelerates at 3 m/s2. How many seconds will it take for the car to catch up to the truck?

Answer 1

Answer:

The car will take 10 s to catch up with the truck.

Explanation:

A motion is uniformly varied rectilinear, abbreviated u.a.r.m (uniformly accelerated rectilinear motion) occurs when the trajectory of the mobile is a straight line and its speed varies the same amount in each unit of time. In this way, the acceleration is constant.  The position in this case is given by the expression:

$x=x0 + v0*t + \frac{1}{2} *a*t^{2}$

In this case, the car accelerates at 3 m/s² starting from rest, that is, the initial speed v0 = 0. Being x0 = 0, then:

$x=\frac{1}{2} *3\frac{m}{s^{2} } *t^{2}$

A movement is rectilinear when a mobile describes a straight path, and it is uniform when its speed is constant in time, since its acceleration is zero. That is, the Uniform Rectilinear Motion (URM) is a straight path, where the velocity is constant and the acceleration is zero. In this case the position at any time t is given by:

x=v*t

In this case the truck travels at 15 m/s through URM using the expression:

x=15 m/s *t

You need to know the time it will take for the car to catch up with the truck, that is, when the position of both will be the same. This is:

$\frac{1}{2} *3\frac{m}{s^{2} } *t^{2}=15 \frac{m}{s} *t$

Solving:

$1.5\frac{m}{s^{2} } *t^{2}=15 \frac{m}{s} *t$

$\frac{t^{2} }{t} =\frac{15\frac{m}{s} }{1.5\frac{m}{s^{2} } }$

$t= 10 s$

The car will take 10 s to catch up with the truck.