Respuesta :
Answer :
(a) The acceleration is, [tex]0.278m/s^2[/tex]
(b) Distance it travels during this time is, 156.225 m
Solution for part (a) :
Formula used :
[tex]a=\frac{v-u}{t}[/tex]
where,
a = acceleration
v = final velocity = [tex]45Km/hr=45\times \frac{5}{18}=12.5m/s[/tex]
u = initial velocity = [tex]30Km/hr=30\times \frac{5}{18}=8.33m/s[/tex]
t = time = 15 s
Now put all the given values in the above formula, we get
[tex]a=\frac{(12.5-8.33)m/s}{15s}=0.278m/s^2[/tex]
Therefore, the acceleration is, [tex]0.278m/s^2[/tex]
Solution for part (b) :
Formula used :
[tex]s=u\times t+\frac{1}{2}a\times t^2[/tex]
where,
s = distance traveled
Now put all the given values in this formula, we get
[tex]s=(8.33m/s)\times (15s)+\frac{1}{2}\times (0.278m/s^2)\times (15s)^2=156.225m[/tex]
Therefore, the distance it travels during this time is, 156.225 m
The acceleration of the train is 0.278 m/s² and the distance traveled by the train is 156.23 m.
The given parameters;
- initial velocity of the train, u = 30 km/h = 8.33 m/s
- final velocity of the train, v = 45 km/h = 12.5 m/s
- time of motion, t = 15 s
The acceleration of the train is calculated by using the kinematic equation as follows;
[tex]a = \frac{\Delta v}{\Delta t}= \frac{12.5 - 8.33}{15} = 0.278 \ m/s^2[/tex]
The distance traveled by the train is calculated using the third kinematic equation as shown below;
v² = u² + 2as
[tex]s = \frac{v^2 - u^2}{2a} \\\\s = \frac{(12.5)^2 - (8.33)^2}{2(0.278)} \\\\s = 156.23 \ m[/tex]
Thus, the acceleration of the train is 0.278 m/s² and the distance traveled by the train is 156.23 m.
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