A motorcycle traveling at a constant speed of 18\text{ m/s} passes a landmark. Exactly 4.0… — JAMB UTME

A motorcycle traveling at a constant speed of $18\text{ m/s}$ passes a landmark. Exactly $4.0\text{ s}$ after passing the landmark, the rider accelerates uniformly at a rate of $2.5\text{ m/s}^2$ until reaching a speed of $28\text{ m/s}$ . What is the total distance, in meters, traveled by the motorcycle from the instant it passed the landmark to the moment it reaches $28\text{ m/s}$ ?

Cevap: 164 m

Cevap

The total distance traveled by the motorcycle is

164\text{ m}

The total distance is obtained by finding the displacement during the

4.0\text{ s}

of constant speed at

18\text{ m/s}

(

72\text{ m}

) and adding the displacement during uniform acceleration from

18\text{ m/s}

28\text{ m/s}

2.5\text{ m/s}^2

(

92\text{ m}

), giving a sum of

164\text{ m}

Adım Adım Çözüm

Calculate the distance covered during the initial constant speed stage.

The distance covered in the first

4.0\text{ s}

72\text{ m}

At a constant velocity

v = 18\text{ m/s}

, the distance

s_1 = v \times t = 18 \times 4.0 = 72\text{ m}

Calculate the distance covered during the accelerated motion stage.

The distance covered while accelerating from

18\text{ m/s}

28\text{ m/s}

92\text{ m}

Using the kinematic equation

v^2 = u^2 + 2as_2

, substitute

u = 18\text{ m/s}

v = 28\text{ m/s}

, and

a = 2.5\text{ m/s}^2

to get

28^2 = 18^2 + 2(2.5)s_2 \implies 784 = 324 + 5s_2 \implies s_2 = 92\text{ m}

Find the total distance traveled.

Total distance

s_{\text{total}} = 164\text{ m}

The total distance is the sum of the distance covered during the constant speed period (

72\text{ m}

) and during uniform acceleration (

92\text{ m}

Anahtar Kavram

Kinematics equations for multi-stage motion combining uniform speed and uniform acceleration.

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