Question

A car going at 30 m/s undergoes an acceleration of 2 m/s^2 for 4 seconds. What is it's final speed? How far did it travel while it was accelerating

Answer 1

In this situation we need to use an kinematics equation. Usually I get questions about what is final velocity not speed.
KINEMATIC EQUATION:
Final velocity= initial velocity + acceleration×time

Substitute factors
Final velocity= 30 + 2(4)
Multiply
Final velocity= 30 + 8
Now add
Final velocity (or final speed)= 38 m/s

Now you want to find how long did it travel, we have to use another kinematics equation
KINEMATIC EQUATION:
Distance= initial velocity×time + (1/2)×acceleration× time squared

Substitute
distance= 30(4) + (1/2)(30)(4^2)
Multiply
distance= 120 + (15)(16)
distance= 120 + 240
distance= 360 m

Answer 2

Explanation:

It is given that,

Initial velocity of the car, u = 30 m/s

Acceleration of the car, a = 2 m/s²

Time taken, t = 4 seconds

(1) We need to find the final velocity of the car. It can be calculated using first equation of motion as :

$v=u+at$

$v=30\ m/s+2\ m/s^2\times 4\ s$

v = 38 m/s

The final speed of the car is 38 m/s.

(2) Let s is the distance travelled by the car. Using third equation of motion as :

$v^2-u^2=2as$

$s=\dfrac{v^2-u^2}{2a}$

$s=\dfrac{(38\ m/s)^2-(30\ m/s)^2}{2\times 2\ m/s^2}$

s = 136 meters

So, while accelerating it will travel 136 meters. Hence, this is the required solution.