## Traveling Train

At maximum acceleration, a train is traveling 60 m/s. Two minutes later, it is going 75 m/s. Calculate the amount of seconds it would take the train to stop when traveling at 35 m/s, if it can decelerate 3 times faster than it can accelerate.

##
__
__**Hint**

**Hint**

$$$v=a_0t+v_0$$$

where
$$v$$
is the final velocity,
$$a_0$$
is constant acceleration,
$$t$$
is time, and
$$v_0$$
is the initial velocity.

##
__
__**Hint 2**

**Hint 2**

Use the above equation to first solve for acceleration, then again for the time.

For constant acceleration, the equation of motion is:

$$$v=a_0t+v_0$$$

where
$$v$$
is the final velocity,
$$a_0$$
is constant acceleration,
$$t$$
is time, and
$$v_0$$
is the initial velocity. First, letâ€™s solve for acceleration from the increasing speed scenario:

$$$75m/s=a_0(2*60sec)+(60m/s)$$$

$$$a_0=\frac{(75m/s-60m/s)}{120sec}=\frac{15m}{120s^2}=0.125\:m/s^2$$$

Because the train can decelerate 3 times faster than it can accelerate, the maximum deceleration is:

$$$a_{decel}=-a_0\times 3= (0.125m/s^2)(3)=-0.375\:m/s^2$$$

In the slowing down scenario, acceleration is still constant, so we can reuse the same equation to solve for the stopping time:

$$$v=a_{decel}(t)+v_0$$$

$$$0m/s=(-0.375m/s^2)(t)+(35m/s)$$$

$$$t=\frac{-35m/s}{-0.375m/s^2}=93.3\:sec$$$

93.3 sec