Question

A cyclist moving towards the right with an acceleration of 4ms-2 at t=0 has covered 5m moving toward the right at 15ms-1 what will be his position at t=2 sec?
a,36m
b,38m
c,41m
d,43m​

Answer 1

Answer:

38 m

Explanation:

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Answer 2

Answer:

Note there appears to be one mistake as velocity is 15m/s. And “moving towards the right” would mean MY Right and not the cyclist’s right !! So let us assume the following:-

1. at t= 0, acceleration is 4m/ss, then he traveled a distance of 5m and finished with a velocity of 15m/s after that first sprint

2. One wants to know where would the cyclist be in 2 seconds FROM THE START at t=0.

Using vv=uu +2as = 15*15 =uu+2.4.5 hence starting velocity/speed =u= sqrt 185

Using v=u+at from the start then the final velocity = 15 +4.2= 23

The distance covered from the start is

(15+23)/2 = average speed times 2 seconds = 38 metres . Adding the initial 5 m mentioned that would bring the 43.

From zero speed, the cyclist must have been on the road for 5.4 seconds accelerating at 4m/ss and covered a total distance of 58.32 metres.

I need to revise the last part. an interesting question where three time periods need to be considered. Starting from zero speed, and the two time periods mentioned in the question.

Your question is garbled.

First you say he has an acceleration of 4 m/s^2 (but don’t tell us his velocity). Then you say that he HAS travelled 5 m with an acceleration of 15 m/s^2 (or did you mean velocity - but then the units are wrong). But you didn’t say what happened before that. I have no idea what he did between t=0 and t=2, nor how you define the zero of position.

So, first decide clearly what it is you know and what you want to know

Explanation:

So the answer is 43

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