A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.42m/s2 and β = 5.00×10−2m/s3 .
Calculate the average velocity of the car for the time interval t=0 to t1 = 2.08s .
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Question

A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.42m/s2 and β = 5.00×10−2m/s3 .
Calculate the average velocity of the car for the time interval t=0 to t1 = 2.08s .
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anonymous · Answer

dude..... α is in m/s2 as acceleration..... i got it......

but what happen to the β which is in m/s3.... which unit is it...... clarity please..........

anonymous · Answer

idk im really confused at thsi question

anonymous · Answer

@harishk see whole term β t3 which is unit of x

anonymous · Answer

x(t)= α t2− β t3 calculate v(t)= dx(t)/dt =x(t)= 2α t− 3β t2
so now calculate v(t) at t=0 and t1 = 2.08s 
now average velocity would be v(average)=  [v(t1 = 2.08s)-v(t=0)] / (2.08-0)

anonymous · Answer

sorry in v(t)= dx(t)/dt =x(t) i wrote x(t) by mistake

anonymous · Answer

I understand the the first 2 steps but what do u have to do in the end sorry

anonymous · Answer

that is gyanu its good

anonymous · Answer

oh @thereyruls sorry i made huge mistake
average velocity=( distance travel by car in time interval t=0 to t1 = 2.08)/time interval
= {x(2.08)-x(0)}/(2.08-0)
where x(2.08) is value of x(t) at t=2.08
in x(t)= α t2− β t3, put t= 2.08 you will get x(2.08)
similarly in x(t)= α t2− β t3, put t=0 you will get x(0)
please tell me if you don't get any thing  and lots of sorry for my mistake.