Question

An antelope moving with constant acceleration covers the distance between two points 70.0 m apart in 6.00 s. Its speed as it passes the second point is 15.0 m/s. What are(a) its speed at the first point and (b) its acceleration

Answer 1

Answer:

(a) The speed at first point is 8.33m/s.

(b) The acceleration of the antelope is 1.11m/s².

Explanation:

We have, from the kinematics equations, that:

$a=\frac{v_2-v_1}{t} \\\\v_2^{2}=v_1^{2}+2ax$

Solving for the acceleration in the second equation, we have:

$a=\frac{v_2-v_1}{t} \\\\a=\frac{v_2^{2} -v_1^{2} }{2x}$

Then, equating the two equations, we obtain:

$\frac{v_2-v_1}{t}=\frac{v_2^{2} -v_1^{2} }{2x}\\\\\frac{2x}{t}=\frac{(v_2-v_1)(v_2+v_1)}{v_2-v_1}\\\\v_1=\frac{2x}{t}-v_2\\\\\implies v_1=\frac{2(70.0m)}{6.00s}-15.0m/s=8.33m/s$

So, the speed at the first point is 8.33m/s (a).

Next, we use the first equation to compute the acceleration:

$a=\frac{15.0m/s-8.33m/s}{6.00s}=1.11m/s^{2}$

In words, the acceleration of the antelope is 1.11m/s² (b).