Question

A motorcyclist started a race from START line and from rest, and accelerated with the acceleration of 2 m/s2 for 12 seconds. Then he increased the acceleration to 5 m/s2 and went with that acceleration for 4 seconds. What was the racer’s position from the START line after 12+4 seconds from the start of the race?
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Answer 1

Answer:

The total distance is 280 [m]

Explanation:

In order to solve this problem we must use the expressions of kinematics. The clue to solve this problem is that the motorcyclist starts from rest, i.e. its initial speed is zero.

$v_{f} =v_{o} +(a*t)$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0

a = acceleration = 2 [m/s²]

t = time = 12 [s]

Vf = 0 + (2*12)

Vf = 24 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

24² = 0 + (2*2*x)

x = 576/(4)

x = 144 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

The other important clue to solve this problem in the second part is that the final velocity is now the initial velocity.

We must calculate the final velocity.

$v_{f}= v_{i} +(a*t)$

Vf = final velocity [m/s]

Vi = initial velocity = 24 [m/s]

a = desacceleration = 5 [m/s²]

t = time = 4 [s]

Vf = 24 + (4*5)

Vf = 44 [m/s]

With this velocity, we can calculate the displacement using the following expression.

$v_{f} ^{2} =v_{o} ^{2} +2*a*x$

where

x = distance traveled [m]

44² = 24² + (2*5*x)

x = 1360/(10)

x = 136 [m]

Note: The positive sign in the equations is because the car is accelerating, it means its velocity is increasing.

Therefore the total distance is Xt = 144 + 136 = 280 [m]