While driving, your car has an initial position of 3.2 m, an initial velocity of -8.4 m/s, and
constant acceleration of 1.1 m/s^2 what is the position of the car at the time t= 1.5 s?

The position of the car at t = 1.5 s is at -8.1625 meters

Explanation:

The initial position of the car is 3.2 meters

The initial velocity is -8.4 m/s

The constant acceleration is 1.1 m/s²

We need to find the final position of the car at the time t = 1.5 seconds

The displacement s = final position - initial position

, where u is the initial velocity, a is the

constant acceleration and t is the time

So we can find the final velocity by using the rule:

final position - initial position =

initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s

Substitute these values in the rule

final position - 3.2 =

final position - 3.2 = -12.6 + 1.2375

final position - 3.2 = -11.3625

add 3.2 for both sides

final position = -8.1625

That means the car is at 8.1625 meters in opposite direction

The position of the car at t = 1.5 s is at -8.1625 meters

no. dot product   =0 when vecs at right angles