Question

An object was launched from a place P in constant speed to hit a target. At the $15$th second it was $1400m$ away from the target and at the 18th second $800m$ away. Find (1) the distance between the place and the target (2) the distance covered by it in $15$ seconds,(3) time taken to hit the target.

Answer 1

Here, P denotes the place from where the object was launched,

A and B denotes the positions of object at ${15}^{th}$ and ${18}^{th}$ second.

$\therefore$ Distance covered by the object between ${15}^{th}$ and ${18}^{th}$ second (i.e. 3 seconds) = $1400m$ - $800m$ = $600m$

So, the speed of the object = $\frac{distance}{time}$ = $\frac{600m}{3s}$ = $200m/s$

Thus, Distance covered in 15 seconds (PA) = speed $\times$ time = $200m/s$ $\times$ $15s$ = $3000m$

Now from figure , distance between place and target = PA + $1400m$ = $3000m$ + $1400m$ = $4400m$

Total time taken to hit the target = $\frac{distance}{speed}$ = $\frac{4400m}{200m/s}$ = $22s$

So,

(1) the distance between the place and the target = $4400m$

(2) the distance covered by it in 1515 seconds = $3000m$

(3) time taken to hit the target = $22s$.