Question

Un carro parte del reposo y acelera uniformemente a 4.0 m / s^2 durante 5.0 s. A continuación, mantiene la velocidad que alcanzó durante 10 s. Luego, disminuye la velocidad a una velocidad constante de 2.0 m / s^2 durante 4.0 s.

Answer 1

Answer:

Average speed of the car = 16.53 m.s

Explanation:

For first 5 seconds:

As the car starts from rest, so the initial speed of the car, u=0

Acceleration, a= 4 m/s^2

So, the final speed (speed at time, t=5 second)

v=u+at

v=0 + 4 x 5 = 20 m/s

Distance covered,

$s=ut+\frac 1 2 a t^2 \\\\s=0+\frac 1 2 \times 4 \times 5^2$

s=50 m

As the car maintained the constant speed of 20 m/s for the next 10 seconds.

Distance traveled in 10 seconds, d=20 x 10 =200 m

For the last 4 seconds:

The car slows down at the rate of 2.0 m / s ^ 2 from the speed of 20 m/s.

As the car slows down, so the acceleration will be negative, i.e a=-2 m/s^2

$s=ut+\frac 12 at^2 \\\\s=20\times 4 +\frac 12 (-2)4^2$

s=80-16=64 m.

Considering all three cases, the total distance covered=

=50+200+64=314 m

Total time= 5+10+4=19 s

So, the averate peed= Totat distance covered / total time taken

=314/19=16.53 m/s