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As a train accelerates away from a station, it reaches a speed of 4.5m/s in 4.8s . If the train's acceleration remains constant, what is its speed after an additional 7.0s has elapsed?
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Hello @guiwuzhe456 ! Welcome to OpenStudy !
Do you the equations of motion ?
Using the equation v=u+at We cn frst find the acceleration of the train as 4.5=0+a*4.8 => a= 4.5/4.8=0.937 m/s^2 Now again using the same equation we can find the final final velocity as v=0+0.937*11.8 => v= ?
v=11.0566
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