A car driving on a straight track accelerates at a rate of 3.2 m/s^2 for 14 sec. If the initial velocity of the car was 5.1 m/s, and its initial position was 0 m, what is its final position?
-385m
32m
385m
242.2m
-242.2m

The final position of the car is 385 m.

Explanation:

We know the formula :

                                   s =  

where,

                                     s = distance

                                     u = initial velocity = 5.1 m/s

                                      t = time taken = 14 sec

                                     a = acceleration = 3.2 m/

Solving:

                         s = 5.1* 14 + * 3.2 *

                      ∴ s = 385 m

Hence the final position of the car is 385 m.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

0.863333

86%

hope this