Abiker pedals with a constant acceleration to reach a velocity of 7.5m/s over 4.5s. during the period of acceleration, the bike's displacement is 19m. what was the initial velocity of the bike?

Given:

v(final velocity)= 7.5 m/s

t(time taken to pedal along)=4.5 s

Displacement (s)=19 m

Now we know that

s=ut+1/2(at^2)

Where s is the displacement measured in m

u is the initial velocity measured in m/sec

a is the acceleration measured in m/s^2.

t is the time taken to cover this distance.

Substituting the given values in the above formula we get

19= 4.5u+1/2(a x 4.5 x 4.5)

20.25 a + 9 u = 38

Now we also know that

v= u + at

Substituting the given values in the above formula we get

7.5= u + 4.5a

Solving for u and a from the above equations we get

u = 0.944m/s

a= 1.45 m/s^2

Thus the initial velocity is 0.944 m/s

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

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