Question

A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are running at full speed, with the gazelle 96.8 m ahead, how long before the cheetah catches its prey? Answer in units of s.
The cheetah can maintain its maximum speed for only 7.5 s. What is the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape? (After 7.5 s the speed of cheetah is less than that of the gazelle.) Answer in units of m.

Answer 1

(1)

Cheetah speed: $v_c = 97.8 km/h=27.2 m/s$

Its position at time t is given by

$S_c (t)= v_c t$ (1)

Gazelle speed: $v_g = 78.2 km/h=21.7 m/s$

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

$S_g(t)=S_0 +v_g t$ (2)

The cheetah reaches the gazelle when $S_c=S_g$. Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

$v_c t=S_0 + v_g t$

$(v_c -v_g t)=S_0$

$t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s$

(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: $S_c = v_c t =(27.2 m/s)(7.5 s)=204 m$

Gazelle: $S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m$

So, the gazelle should be ahead of the cheetah of at least

$d=S_c -S_g =204 m-162.8 m=41.2 m$

Answer 2

The cheetah catches the prey $\fbox{17.78 s}$ before the gazelle.

The minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is $\fbox{40.83 m}$.

Further Explanation:

The speed is the rate of change of the distance and the relative speed is the speed of the one object in respect of another object. If two body moves in same direction then the relative speed in respect of the one body is given by subtract of both speed and it they are moving in opposite direction the relative speed in respect of the one body is given by sum of both speeds.

Given:

The maximum speed of the cheetah is $97.8 km/h$.

The maximum speed of the gazelle is $78.2 km/h$.

The distance between the starting points of cheetah and the gazelle is $96.8 m$.

Concept:

The expression for the position of the cheetah at some time t is:

${S_1}={v_c}t$                                            …… (1)

Here, ${S_1}$the position of the cheetah at some time $t$, ${v_c}$ is the speed of the cheetah and ${S_1}$ is the distance covered at time $t$.

The expression for the position of the gazelle at some time $t$ is:

${S_2}={S_o}+{v_g}t$                                              …… (2)

Here, ${S_2}$ the position of the gazelle at some time $t$, ${v_g}$ is the speed of the gazelle, ${S_o}$ is the distance between the starting points of cheetah and the gazelle.

At the time when the cheetah reached the gazelle then ${S_1}={S_2}$.

Equate equation (1) and (2).

$\begin{aligned}{v_c}t&={S_o}+{v_g}t\hfill\\{v_c}t-{v_g}t&={S_o}\hfill\\t\left( {{v_c} - {v_g}}\right)&={S_o}\hfill\\t&=\frac{{{S_o}}}{{\left( {{v_c} - {v_g}} \right)}}\hfill\\\end{aligned}$

Substitute $97.8 km/h$ for ${v_c}$, $78.2 km/h$ for ${v_g}$ and $96.8 m$ for ${S_o}$ in the above equation.

$\begin{aligned}t&=\frac{{96.8{\text{ m}}}}{{\left( {97.8 - 78.2} \right){\text{km/h}}\left({\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)}}\\&=\frac{{96.8{\text{ m}}}}{{{\text{19}}{\text{.6 m/s}}\left( {\frac{5}{{18}}}\right)}}\\&=17.78{\text{s}}\\\end{aligned}$

Therefore, the cheetah catches the prey $17.78 s$ before the gazelle.

To calculate the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape we should subtract the distance covered by the cheetah and gazelle in $7.5 s$.

The expression for the distance covered is:

$S = vt$

Here, $S$ is the distance covered, $v$ is the speed and $t$ is the travel time.

Substitute $97.8 km/h$ for $v$ and $7.5 s$ for $t$ in the above equation.

$\begin{aligned}S&=\left( {97.8{\text{ km/h}}} \right)\left( {\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)\left( {7.5{\text{ s}}} \right)\\&=\left( {97.8{\text{ m/s}}} \right)\left( {\frac{5}{{18}}} \right)\left( {7.5{\text{ s}}} \right)\\&=203.75{\text{m}}\\\end{aligned}$

Therefore, the distance covered by the cheetah in $7.5 s$ is $203.75 m$.

Substitute $78.2 km/h$ for $v$ and $7.5 s$ for $t$ in the above equation.

$\begin{aligned}S&=\left( {78.2{\text{ km/h}}} \right)\left( {\frac{{1000{\text{ m}}}}{{1{\text{ km}}}}} \right)\left( {\frac{{1{\text{ h}}}}{{{\text{3600 s}}}}} \right)\left( {7.5{\text{ s}}} \right)\\&=\left( {78.2{\text{ m/s}}} \right)\left( {\frac{5}{{18}}} \right)\left( {7.5{\text{ s}}} \right)\\&=162.92{\text{ m}}\\\end{aligned}$

Therefore, the distance covered by gazelle in $7.5 s$ is $162.92 m$.

The minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is:

$\begin{aligned}d&=\left( {203.75 - 162.92} \right){\text{ m}}\\&=40.83{\text{ m}}\\\end{aligned}$

Therefore, the minimum distance the gazelle must be ahead of the cheetah to have a chance of escape is $\fbox{40.83 m}$.

Learn more:

1.  Conservation of momentum brainly.com/question/7031524.

2.  Motion under friction brainly.com/question/11023695.

3. Motion under force brainly.com/question/4033012.

Answer Details:

Grade: High school

Subject: Physics

Chapter: Kinematics

Keywords:

Cheetah, 97.8 km/h, gazelle, 78.2 km/h, full speed, 96.8 m. ahead, catches, prey, 7.5 s, minimum, ahead, escape, answer, unit, s, m, 40.83 m.