Question

A car starts from rest with a constant acceleration of $2m.{s^{ - 2}}$ . Determine: (a) the speed at the end of fifth second (b) the distance covered in five seconds (c) the speed after covering $100m$ ?

Answer 1

As we know that when any object starts moving from the rest then its initial speed is $0m.{s^{ - 1}}$ . Here, in this question, we will apply three various formulas to find the required speed and the distance.

Formula-used: Formula of speed when the acceleration and time is given- $v = u + at$
And, the formula of distance: $S = ut + \dfrac{1}{2}a{t^2}$ ,and
The formula of speed when the acceleration and speed is given: ${v^2} = {u^2} + 2as$ .

Complete step by step solution:
As per the question, if the car starts from rest that means the initial velocity of a car is zero:
$u = 0m.{s^{ - 1}}$
And, the constant acceleration is given as, $a = 2m.{s^{ - 2}}$ .
(a) We have to find the speed at the end of fifth second, that means the time taken to achieve the speed:
$\because t = 5\,\sec$
Now, to find the speed or the final velocity with respect to the given time, we will apply the formula of speed in the terms of given acceleration and the time:
$\therefore v = u + at$
where, $v$ is the final velocity or the speed at the end of fifth second,
$u$ is the initial velocity, $u = 0m.{s^{ - 1}}$ ,
$a$ is the given constant acceleration, $a = 2m.{s^{ - 2}}$ and
$t$ is the time at which the speed will be achieved.
$\Rightarrow v = 0 + 2 \times 5 \\\ \Rightarrow v = 10m.{s^{ - 1}} \\\$
Hence, the speed at the end of the fifth second of a car is $10m.{s^{ - 1}}$ .
(b) Now, to find the distance covered in five seconds, we will apply the formula of distance in the terms of velocity, acceleration and the time taken to cover the distance:
$\therefore S = ut + \dfrac{1}{2}a{t^2}$
where, $S$ is the required distance that is covered in five seconds,
$u$ is the initial velocity of a car,
$a$ is the given constant acceleration, $a = 2m.{s^{ - 2}}$ and
$t$ is the time taken to cover the distance.
$\Rightarrow S = 0 \times 5 + \dfrac{1}{2} \times 2 \times {5^2} \\\ \Rightarrow S = 0 + 25 = 25m \\\$
Hence, the distance covered in five seconds is $25metre$ .
(c) Now, according to the question, if the distance will be $100m$ , then we have to find the speed:
So, we will apply the formula to find the speed in the terms of acceleration and distance:
$\therefore {v^2} = {u^2} + 2as$
where, $v$ is the required speed which we have to find,
$u$ is the initial velocity as the car starts from rest, and
$s$is the new distance which is $100m$ .
$\Rightarrow {v^2} = {0^2} + 2.2.100 \\\ \Rightarrow v = \sqrt {400} = 20m.{s^{ - 1}} \\\$
Hence, the speed after covering $100m$ is $20m.{s^{ - 1}}$ .

Note:
The following all used formulas are the equations of motion. Equations of motion are mathematical expressions that describe the behaviour of a physical system as a function of time. More specifically, equations of motion are a set of mathematical functions that describe the behaviour of a physical system in terms of dynamic variables. Typically, these variables are spatial coordinates and time, but they may also incorporate momentum components. Generalized coordinates are the most versatile option, as they can be any convenient physical system variable.