Answer to Question #100329 in Mechanics | Relativity for EIRRA EIRWANI

Question #100329
An electric train starting from rest attains a maximum
speed of 200 kmph in 20 seconds. Determine :
(i) Its acceleration assuming it to be uniform
(ii) Distance covered during this time period
(iii) Its velocity 15 seconds after staring from rest
1
Expert's answer
2019-12-16T10:55:09-0500

From the formula for univariate movement

v=v0+atv= v_0+a \cdot t

where by the condition of the problem v=200km/h=55.56m/sv= 200 {km/h}=55.56{m/s} v0=0v_0=0

1)define acceleration

a=vv0t=55.56020=2.778[m/s2]a=\frac{v-v_0}{t}=\frac{55.56-0}{20}=2.778[m/{s^2}]

2) Distance traveled is

s=at22=2.7782022=555.6ms= \frac{a \cdot t^2}{2}=\frac{2.778 \cdot 20^2}{2}=555.6m

3)Its speed after 15 seconds from the start of movement

v=at1=2.77815=41.67m/sv=a \cdot t_1=2.778 \cdot 15=41.67{m/s}


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