Two motorcycles are traveling due east with different velocities. however, 3.49 seconds later, they have the same velocity. during this 3.49-second interval, motorcycle a has an average acceleration of 2.52 m/s2 due east, while motorcycle b has an average acceleration of 17.5 m/s2 due east. (a) by how much did the speeds differ at the beginning of the 3.49-second interval, and (b) which motorcycle was moving faster?

(a) The speeds differ by 52.2802 m/s at the beginning of 3.49 s interval

(b) The first motorcycle was moving faster

Explanation:

The time of the 2 motorcycles is 3.49 seconds

The final velocities of them are equal

The travelling on the same direction (due east)

The acceleration of the first one is 2.52 m/s²

The acceleration of the second one is 17.5 m/s²

(a)

we need to find the difference between their speeds at the beginning

of the 3.49 s interval

Assume that their final velocity is v

→ v = + t

→ v = + t

→ = 2.52 m/s² , = 17.5 m/s² , t = 3.49 s

- Substitute these value in the equations above

→ v = + (3.49)

  v = + 8.7948 ⇒ (1)

→ v = + (3.49)

  v = + 61.075 ⇒ (2)

Equate equations (1) and (2)

→ + 8.7948 = + 61.075

Subtract 8.7948 from both sides

→ = + 52.2802

Subtract from both sides

→ - = 52.2802 m/s

The speeds differ by 52.2802 m/s at the beginning of 3.49 s interval

(b)

The difference between the speed of the 1st motorcycle and the speed

of the 2nd motorcycle is positive then the 1st motorcycle was moving

faster

The first motorcycle was moving faster

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