Going 8.4m/s on a bike, go across 7.2m sand, speed is 6.4m/s when out. bike's accel in sand? time to cross?

Question

Going 8.4m/s on a bike, go across 7.2m sand, speed is 6.4m/s when out. bike's accel in sand? time to cross?

Answer 1

Note that ^ means there is an exponent for example 2^2 means 2 squared

Note: V0 is initial velocity, Δx is total distance, Vf is final velocity, and a is acceleration.

Given: V0= 8.4 m/s, Δx= 7.2 m, Vf=6.4 m/s, a= ?

So, we use the equation Vf^2 = V0^2 + 2a (Δx)

which is then Vf^2- V0^2=2a (Δx). So, then we have

(6.4 m/s)^2-(8.4 m/s)^2=2a(7.2m)

-29.6 m^2/s^2=2a(7.2m)

So (-29.6 m^2/s^2)/(2*7.2 m)=a

a= - 2.1 m/s^2

the time is got by taking Vf=V0+at

this so 6.4 m/s=8.4 m/s + (-2.1 m/s^2)t

So, -2.0 m/s=(-2.1 m/s^2)t

So, t=0.95 s

Answer 2

a)  Find acceleration of bike in sand.

     Solution:

     2ad = Vf^2 - Vi^2

         a = (Vf^2 - Vi^2)/(2d)

         a = [(6.4m/s)^2 - (8.4m/s)^2]/[(2)(7.2m)]

         a = - 2.06 m/s^2    ANSWER

b)  Find time to cross.

     Solution:

     d = vt, where v = average velocity, and t = time to cross, and

           d = 7.2 m

     v = (Vf + Vi)/2 = (6.4 m/s + 8.4 m/s)/2 = 7.4 m/s

     Therefore,

     t = d/v = 7.2 m/7.4 m/s

     t = 0.97 s    ANSWER

Hope this helps.

teddy boy