Question

a cyclist driving at 5m/s picks a velocity of 10m/s over distance of 50 m calculate -1) Acceler ,2) time in which cyclist picks up t t he above velocity l.

Answer 1

INITIAL VELOCITY = 5 M/S

FINAL VELOCITY = 10 M/S

DISTANCE COVERED = 50 M

USING 3RD EQUATION OF UNIFORMLY ACCELERATED MOTION,

${v}^{2} = {u}^{2} + 2as \\ {10}^{2} = {5}^{2} + 2 \times a \times 50 \\ 100 = 25 + 100a \\ 100a = 75 \\ a = \frac{75}{100} \\ a =0. 75 \: m {s}^{ - 2}$

USING 1ST EQUATION OF UNIFORMLY ACCELERATED MOTION,

$v = u + at \\ 10 = 5 + (0.75 \times t) \\ 5 = 0.75t \\ 0.75t = 5 \\ t = \frac{5}{0.75} \\ t = \frac{500}{75} \\ t = \frac{20}{3} \\ t = 6.66...seconds$
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Answer 2

Answer:

Explanation:answer is 0.75 ms^-2

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