A particle moves along the x axis with velocity at time tgiven by v= 5t^2 + 2t.

Question

A particle moves along the x axis with velocity at time tgiven by v= 5t^2 + 2t. Find: (a) the distance it moves in the 2nd second (b) the distance it moves in the 4th second. I know you have to integrate the velocity to finddisplacement, i call this x . I got x = 5t^3/3 + t^2 butcan't seem to get the answers for distance required right bysubstituting, the answers apparently are: (a) 14.7m (b) 68.7m A particle moves along the x axis with velocity at time tgiven by v= 5t^2 + 2t. Find: (a) the distance it moves in the 2nd second (b) the distance it moves in the 4th second. I know you have to integrate the velocity to finddisplacement, i call this x . I got x = 5t^3/3 + t^2 butcan't seem to get the answers for distance required right bysubstituting, the answers apparently are: (a) 14.7m (b) 68.7m

Explanation / Answer

What i think you are doing wrong is just intepretation of thequestion. The integration is the correct one, but what you are supposed tocalculate is the distance it moves during the 2nd second and not in2 seconds. So for a): x=(5*2^3)/3 + 2^2 - ((5*1^3)/3 + 1^2) and for b) x=(5*4^3)/3 + 4^2 - ((5*3^3)/3 + 3^2) Basically you take the total amount moved during four seconds andthen remove the amount moved during the first three, leaving justthe distance during the 4th second

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