While entering a freeway, a car accelerates from rest at a rate of 2.1 m/s^2 for 14.5 s. how far does the car travel in those 14.5 s, in meters?
what is the car's final velocity, in meters per second?

220.8 m

30.5 m/s

Explanation:

a = acceleration of the car = 2.1 m/s²

t = time of travel = 14.5 s

x = displacement of the car

v₀ = initial velocity of the car = 0 m/s

Displacement of the car is given as

x = v₀ t + (0.5) a t²

Inserting the values

x = (0) (14.5) + (0.5) (2.1) (14.5)²

x = 220.8 m

v = final velocity of the car

Final velocity of the car is given as

v = v₀ + at

v = 0 + (2.1) (14.5)

v = 30.5 m/s

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.

when jerome paddle in flowing river towards east the net speed of jerome will increase

so the net speed of jerome will be given as

now the speed of paul will be same as it is given as he is running on the ground

so

so here we can say since the speed of jerome is more so he will win the race

now as we know the distance so the time difference is given as

so jerome will reach first and the time difference will be 203.55 s or 3.4 min