Question

child slide down a hill with an acceleration of 2.9 m/s^2 if she start with an initial velocity of 2.3 how far does she travel in 5.0 second

Answer 1

Answer:

 Distance traveled in 5 seconds = 47.75 meter.

Explanation:

 We have equation of motion , $s= ut+\frac{1}{2} at^2$, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 In this the velocity of body = 2.3 m/s, acceleration = 2.9 $m/s^2$, we need to calculate displacement when t = 5 seconds.

  $s= 2.3*5+\frac{1}{2} *2.9*5^2=47.75m$

So, distance traveled in 5 seconds = 47.75 meter.