A stationary climber accidentally releases a torch. The torch falls for 1.40 s before reaching a ledge below.
Ignore air resistance and use g=10 m/s2g=10\ \mathrm{m/s^2}g=10 m/s2.
Use the equations a=v−ut\displaystyle a=\frac{v-u}{t}a=tv−u and v2−u2=2axv^2-u^2=2axv2−u2=2ax.
What is the distance travelled by the torch between 0.60 s and 1.40 s after its release?
8.0 m8.0\ \mathrm{m}8.0 m
4.0 m4.0\ \mathrm{m}4.0 m
3.2 m3.2\ \mathrm{m}3.2 m
9.8 m9.8\ \mathrm{m}9.8 m
96 exam-style questions on Edexcel GCSE Physics 2.2 Speed, velocity and acceleration, covering 2.2.1 Speed and distance–time graphs, 2.2.2 Acceleration and the uniform-acceleration equation, 2.2.3 Velocity–time graphs, and 2.2.4 Measuring speed, typical speeds and free-fall acceleration. Each one has a worked solution and a mark scheme showing where the marks go.