A cyclist is moving towards the right with an acceleration of 4m/sec2. At t=0, he has traveled 5m moving towards the right at 15 m/sec. What will be his position at t = 2 seconds?

Explanation

The final position of the cyclist at t=2 seconds will be 43 m Hence, D is the correct option. Explanation: As per the data given in the question, We need to find the position of the cyclist at the time interval t=2 sec. As per the question, It is given that, The value of the acceleration =+4 m/sec. At t=0,x=5m. V=15m/sec. Taking + as the right direction Now, for determining the value of v, we will use the first equation of motion we can say Vf=Vi+a×t. So, Vf=15+4*2=23. Vf=23m/s .Now using the third equation of motion, we can say: Vf²-Vi²=2as So, the value of S will be 23²-15² = 529-225 = 304 = 38 S= 38m Now we know that from all the three equations of motion, we actually get the delta s or say change of distance so here delta s=5−Xf. Here Xf is the final position of the particle, Putting values of s in s=5−Xf we get, Xf=43 m. We also need to understand in questions related to the equation of motion if the value of time is given then we prefer to use the first two equations as they involve the variable of time otherwise we will use the third equation, making it simple for similar problems.