Question

It takes an object 2.99 s to accelerate uniformly from 4 m-s-1 to 11.3 m-s-1A) Calculate the magnitude of the object's acceleration (in m-s2).B) How long (in s) would it take to change the object's speed from 8.2 m-s-1 to 13.6 m.s-1?

Answer 1

Given data:

* The time is taken by the object in the given case is t = 2.99 s.

* The initial velocity of the object is u = 4 m/s.

* The final velocity of the object is v = 11.3 m/s.

Solution:

(A). The magnitude of the acceleration of an object is,

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

Substituting the known values,

$\begin{gathered} a=\frac{11.3-4}{2.99} \\ a=2.44ms^{-2} \end{gathered}$

Thus, the magnitude of the acceleration is 2.44 meters per second squared.

(B). The time is taken by the object to change its velocity from 8.2 m/s to 13.6 m/s with the same acceleration is,

$\begin{gathered} a=\frac{final\text{ velocity - initial velocity}}{t} \\ t=\frac{final\text{ velocity - initial velocity}}{a} \end{gathered}$

Substituting the known values,

$\begin{gathered} t=\frac{13.6-8.2}{2.44} \\ t=2.21\text{ s} \end{gathered}$

Thus, the time taken by the object to change its speed from 8.2 m/s to 13.6 m/s is 2.21 seconds.