Question

A body moves with uniform velocity of $u = 7\,m/s$ from $t = 0$ to $t = 1.5\,{\text{s}}$. It starts moving with an acceleration of $10\,m{s^2}$. The distance between $t = 0$to$t = 3s$will be:
A. $47.75 m$
B. $32.25 m$
C. $16.75 m$
D. $27.50 m$

Answer 1

Hint:We will use the laws of motion to solve this question. We can calculate the distance travelled in the first 1.5 seconds and the distance travelled in the last 1.5 seconds. And then add both the distances to get the total distance.

Formula used:
Laws of motion:
$s = ut + \dfrac{1}{2}a{t^2}$
Here, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Complete step by step answer:
The body is said to be in its uniform velocity only if the body travels an equal distance in equal intervals of time period along a particular direction. We know that from \[t = 0\] to \[t = 1.5\], the body is moving with the uniform velocity, $u =7\,m/s$.

So in the formula:
$s = ut + \dfrac{1}{2}a{t^2}$
We can put $a = 0$as for constant velocity, acceleration is zero.
$ \Rightarrow s = 7 \times 1.5$
$ \Rightarrow s = 10.5m$ . . . (1)
Now after, \[t = 1.5\] the body gets an acceleration of $10\,{\text{m/}}{{\text{s}}^{\text{2}}}$
So, we can again use the formula:
$s = ut + \dfrac{1}{2}a{t^2}$
To get the distance covered by the body in the last \[t = 1.5\].

Now by substituting the values in the above formula, we have,
$s = (1.5 \times 7) + \dfrac{1}{2} \times 10 \times {(1.5)^2}$
$ \Rightarrow s = (1.5) + (5 \times 2.25)$
$ \Rightarrow s = 10.5 + 11.25$
$ \Rightarrow s = 21.75m$ . . . (2)
The total distance covered by the body = sum of the distances covered by the body in the first 1.5 sec and the last 1.5 sec.
So by adding equation (1) and (2), we get,
Total distance covered $ = 10.5m + 21.75m$
$\therefore$Total distance covered = $32.25m$

Hence, from the above explanation, the correct option is B.

Note:You cannot solve this question by directly, putting \[t = 3\,{\text{s}}\] in the equation of displacement as the acceleration changes, and the equations of motion, $s = ut + \dfrac{1}{2}a{t^2}$is used for the constant acceleration in that time period.