Question

g A top-fuel dragster starts from rest and has a constant acceleration of 44.0 m/s2. What are (a) the final velocity of the dragster at the end of 2.1 s, (b) the final velocity of the dragster at the end of of twice this time, or 4.2 s, (c) the displacement of the dragster at the end of 2.1 s, and (d) the displacement of the dragster at the end of twice this time, or 4.2 s?

Answer 1

The dragster's velocity v at time t with constant acceleration a is

`v=at`

since it starts at rest.

After 2.1 s, it will attain a velocity of

`v=\left(44.0(\rm m)/(\mathrm s^2)\right)(2.1\,\mathrm s)`

or 92.4 m/s.

Doubling the time would double the final velocity,

`v=a(2t)=2at`

so the velocity would be twice the previous one, 184.8 m/s.

The dragster undergoes a displacement x after time t with acceleration a of

`x=\frac12at^2`

if we take the starting line to be the origin.

After 2.1 s, it will have moved

`x=\frac12\left(44.0(\rm m)/(\mathrm s^2)\right)(2.1\,\mathrm s)^2`

or 88 m.

Doubling the time has the effect of quadrupling the displacement, since

`x=\frac12a(2t)^2=4\left(\frac12at^2\right)`

so after 4.2 s it will have moved 352 m.