Question

A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 25.4 m. Determine the acceleration of the bike.

Answer 1

using this kinematic equation:
\[ V{_F}^2 = V_{I}^{2} + 2a(X_f-X_i) \]
where \[X_f-X_i\] is the distance covered.
\[(7.10)^2= 0^2 + 2a(25.4)\]
\[(7.10)^2/(2*25.4)=a\]
\[a=0.99232 m/s^2\]

Answer 2

\[_{V2}^{2}-_{V1}^{2}=2a \Delta x\]
\[7.10^{2 }-0=2a*25.4\]
a=0.992

Answer 3

u=0, v=7.10m/s, s=25.4m

v^2= u^2 + 2as (USING 3rd LAW OF MOTION)

7.10*7.10 =0 + 2*a*25.4

a= 0.992 m/s^2