As a train accelerates from a station, it reaches a speed of 4.7m/s in 5.0s. If a train acceleration remains constant what is its speed after an additional 6.0s has elapsed?

You will have to assume that it started at speed = 0 at t = 0. The acceleration rate is therefore

a = 4.7/5.0 = 0.94 m/s^2

After 6.0 MORE seconds at the same acceleration rate, the speed will be
0.94 * 11 = 10.34 m/s

11 seconds is the total time from the start of motion.

six seconds is added to the original five seconds, time =11s

To calculate the final speed of the train after an additional 6.0 seconds has elapsed, we can use the equation for constant acceleration:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given:
Initial velocity, u = 4.7 m/s
Time, t = 5.0 s

Using the equation above, we can calculate the acceleration:

v = u + at
4.7 = u + (a)(5.0)
4.7 = u + 5a
Substituting the value of u:
4.7 = 4.7 + 5a
0 = 5a
a = 0

Since the acceleration is zero, it means that the train is moving with a constant speed of 4.7 m/s.

To find the final speed after an additional 6.0 seconds have elapsed, we can directly use the initial speed:

v = u
v = 4.7 m/s

Therefore, the train's speed after an additional 6.0 seconds is still 4.7 m/s.

To solve this problem, we can use the equation of motion:

v = u + at

where:
v = final velocity (unknown)
u = initial velocity (4.7 m/s)
a = acceleration (unknown)
t = time (11.0s = 5.0s + 6.0s)

We need to find the final velocity (v). Since the train's acceleration is constant, we can assume that the acceleration remains the same. Therefore, we can use the first time interval to find the acceleration.

Using the first time interval:

v = u + at
4.7 m/s = 0 + a * 5.0s

We need to solve for the acceleration (a):

a = (4.7 m/s) / 5.0s
a = 0.94 m/s^2

Now, with the acceleration known, we can find the final velocity using the second time interval:

v = u + at
v = 4.7 m/s + (0.94 m/s^2) * 6.0s

Calculating:

v = 4.7 m/s + 5.64 m/s
v = 10.34 m/s

Therefore, the train's speed after an additional 6.0 seconds has elapsed is 10.34 m/s.

drwls, so where did the 11 come from that you multiplied with 0.94?