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The acceleration of a bus is given by ax(t)=αt, where α = 1.28m/s3 is a constant. Part A If the bus's velocity at time t1 = 1.13s is 5.09m/s , what is its velocity at time t2 = 2.02s ? If the bus's position at time t1 = 1.13s is 5.92m , what is its position at time t2 = 2.02s ?
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a?(t) = Î±t

Î± = 1.28m/sï¿½ = accel.

dv/dt = a?(t) =at or dv = atdt

$\int$ dv = $\int$ ?tdt => v = at2 / 2 + C

also bus's velocity at time t1 = 1.13s is 4.96m/s

=> 4.96 = a1.13ï¿½ / 2 + C = 1.28(1.2769) / 2 + C = 0.817216 + C

=> C = 4.96 - 0.817216 = 4.142784 m/s

therefore v = at2 / 2 + 4.142784

at t = 2.02 => v = 1.28(2.02)2/2 + 4.142784 = 6.75424 m/s

ii) v = ds/dt = at2 / 2 + 4.142784

similarly integrating => s = ?tï¿½ / 6 + 4.142784t + D

bus's position at time t1 = 1.13s is 5.92m

=> 5.92 = 1.28(1.13)3 / 6 + 4.142784(1.13) + D

or 5.92 = 4.98882+ D or D = 0.931175 m

therefore s = at3 / 6 + 4.142784t + 0.931175

at t = 2.02 s => s = 1.28(2.02)3 / 6 + 4.142784(2.02) + 0.931175

= 11.05797868 ~= 11.0579 m

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