Question

a cyclist is moving with a speed of 14m/s^-1 .he starts accelerating with a rate of 6ms^-2 and aquires the speed of 18ms^-1 . calculate what distance did he move in aquiring that speed?
Please solve this with full explanation​

Answer 1

Given:

• Initial velocity of cyclist, u = 14 ms⁻¹
• Acceleration rate of cyclist, a = 6 ms⁻²
• Final velocity acquired by cyclist, v = 18 ms⁻¹

To find:

• Distance covered by cyclist for acquiring final velocity from initial, s =?

Formula required:

• Third equation of motion

      2 a s = v² - u²

[ Where a is acceleration, s is distance covered, v is final velocity, u is initial velocity ]

Calculation:

Using third equation of motion

→ 2 a s = v² - u²

→ 2 ( 6 ) s = ( 18 )² - ( 14 )²

→ 12 s = 324 - 196

→ 12 s = 128

→ s = 128 / 12

→ s = 10.67 metres

Therefore,

• Cyclist will cover a distance of approximately 10.67 metres.