Question:

A car is traveling down a straight road. At time t=0, the car has an initial velocity of 20 m/s. The car experiences a constant acceleration of 2 m/s^2 for 5 seconds. After 5 seconds, the car's acceleration decreases to 1 m/s^2 for the next 10 seconds. Calculate the car's final velocity after 15 seconds.

Answer:

To solve this problem, we will use the following kinematic equation:

Where: $v_f$ = final velocity $v_i$ = initial velocity $a$ = acceleration $t$ = time

First, let's calculate the car's velocity after the first 5 seconds of constant acceleration:

Using the given values, $v_i = 20 \, \text{m/s}$ $a = 2 \, \text{m/s}^2$ $t = 5 \, \text{s}$

We use the kinematic equation: [v_f = v_i + at] [v_f = 20 , \text{m/s} + 2 , \text{m/s}^2 * 5 , \text{s}] [v_f = 20 , \text{m/s} + 10 , \text{m/s}] [v_f = 30 , \text{m/s}]

So, after 5 seconds of constant acceleration, the car's velocity is 30 m/s.

Next, let's calculate the car's velocity after the next 10 seconds of constant acceleration:

Using the given values, $v_i = 30 \, \text{m/s}$ $a = 1 \, \text{m/s}^2$ $t = 10 \, \text{s}$

We use the kinematic equation: [v_f = v_i + at] [v_f = 30 , \text{m/s} + 1 , \text{m/s}^2 * 10 , \text{s}] [v_f = 30 , \text{m/s} + 10 , \text{m/s}] [v_f = 40 , \text{m/s}]

So, after 15 seconds, the car's final velocity is 40 m/s.

Therefore, the car's final velocity after 15 seconds is 40 m/s.