Cat29

**1. Homework Statement**

A train has a length of 118 m and starts from rest with a constant acceleration at time

*t*= 0 s. At this instant, a car just reaches the end of the train. The car is moving with a constant velocity. At a time

*t*= 14.0 s, the car just reaches the front of the train. Ultimately, however, the train pulls ahead of the car, and at time

*t*= 34.6 s, the car is again at the rear of the train. Find the magnitudes of

**(a)**the car's velocity and

**(b)**the train's acceleration.

**2. Homework Equations**

x = vt

x = ut + 0.5at^2

**3. The Attempt at a Solution**

At t = 12s:

distance x = 0.5a14^2 = 98a

x + 118 = v14

14v - 118 = 98a

14v = 98a + 118

v = 7a + 8.43

At t = 34.6s:

distance y = 0.5a34.6^2 = 598.58a

y = v34.6

34.6v = 598.58a

v = 17.3a

Combining the two:

17.3a = 7a + 8.43

10.3a = 8.43

(a) v = 14.2 m/s

(b) a = 0.818 m/s^2

At t = 12s:

distance x = 0.5a14^2 = 98a

x + 118 = v14

14v - 118 = 98a

14v = 98a + 118

v = 7a + 8.43

At t = 34.6s:

distance y = 0.5a34.6^2 = 598.58a

y = v34.6

34.6v = 598.58a

v = 17.3a

Combining the two:

17.3a = 7a + 8.43

10.3a = 8.43

(a) v = 14.2 m/s

(b) a = 0.818 m/s^2

Is this correct? If it isn't, where did I go wrong? Thanks.