Question

In January 2004, NASA landed exploration vehicles on Mars. Part of the descent consisted of the following stages:

Answer 1

Acceleration is given by:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

Let's apply the formula to the different parts of the problem:

A) $-20.5 m/s^2$

Let's convert the quantities into SI units first:

$u = 19300 km/h \cdot \frac{1000 m/km}{3600 s/h} = 5361.1 m/s$

$v=1600 km/h \cdot \frac{1000 m/km}{3600 s/h} =444.4 m/s$

t = 4.0 min = 240 s

So the acceleration is

$a=\frac{444.4 m/s-5361.1 m/s}{240 s}=-20.5 m/s^2$

B) $-3.8 m/s^2$

As before, let's convert the quantities into SI units first:

$u = 444.4 m/s$

$v=321 km/h \cdot \frac{1000 m/km}{3600 s/h} =89.2 m/s$

t = 94 s

So the acceleration is

$a=\frac{89.2 m/s - 444.4 m/s}{94 s}=-3.8 m/s^2$

C) $-53.0 m/s^2$

For this part we have to use a different formula:

$v^2 - u^2 = 2ad$

where we have

v = 0 is the final velocity

u = 89.2 m/s is the initial velocity

a is the acceleration

d = 75 m is the distance covered

Solving for a, we find

$a=\frac{v^2-u^2}{2d}=\frac{0^2-(89.2 m/s)^2}{2(75 m)}=-53.0 m/s^2$