Question

A 873-kg (1930-lb) dragster, starting from rest completes a 405.0-m (0.2531-mile) run in 4.913 s. If the car had a constant acceleration, what would be its acceleration and final velocity?

Answer 1

Weight of dragster = 873 kg

Distance traveled = 403.7 m

Time of travel = 4.920 s

Using the kinematic equation;

D=vt+1/2at^2

Answer 2

Acceleration: approximately
`33.56\; {\rm m\cdot s^(-2)}`.

Final velocity: approximately
`164.9\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

Let
`x` denote displacement. Let
`a` denote the acceleration of the vehicle. Let
`t` denote the duration of the acceleration. Let
`u` denote the initial velocity of the vehicle.

In this question, it is given that
`x = 405.0\; {\rm m}`,
`t = 4.913\; {\rm s}`, and
`u = 0\; {\rm m\cdot s^(-1)}` since the vehicle started from rest. Acceleration
`a` is to be found.

Since acceleration
`a` is constant, apply the SUVAT equation
`x = (1/2)\, a\, t^(2) + u\, t` and solve for acceleration
`a`.

Note that since
`u = 0\; {\rm m\cdot s^(-1)}`, the SUVAT equation becomes
`x = (1/2)\, a\, t^(2)`.

Rearrange and solve this equation for
`a`:

`\begin{aligned}a &= (2\, x)/(t^(2)) \\ &= \frac{2 \, (405.0\; {\rm m})}{(4.913\; {\rm s})^(2)} \\ &\approx 33.558\; {\rm m\cdot s^(-2)}\end{aligned}`.

Let
`v` denote the velocity of the vehicle after the acceleration. Apply the SUVAT equation
`v = u + a\, t` to find this velocity:

`\begin{aligned}v &= u + a\, t \\ &= (0\; {\rm m\cdot s^(-1)}) + (33.558\; {\rm m\cdot s^(-2)})\, (4.913\; {\rm s}) \\ &\approx 164.9\; {\rm m\cdot s^(-1)}\end{aligned}`.