Explanations & Calculations
- At t=0 he has passed the signpost 5m & heading east at 15ms−1.
- It is quite straight forward to calculate the displacement of him at a given time on his travel (under the given acceleration) here onwards, using the equation S=x0+u0t+21at2 which is meant for a motion under constant acceleration.
- Note that the S is the total displacement measured eastwards from the signpost. East→+
- Therefore,
SS=5m+15ms−1(t)+21(4ms−2)(t2)=5+15t+2t2⋯(1)
- Then the position at 2s is
St=2=5+15(2s)+2(2s)2=43m
- Velocity at a given time can be calculated using the equation V=u+at (Derivating the equation (1) also yeilds the same relationship).
VVVt=2=15ms−1+4ms−2(t)=15+4t=15+4×2s=23ms−1
- Using the same equation, the time when the velocity becomes 25ms−1 could be found,
25ms−1t=15ms−1+4t=2.5s
- Plugging this in equation (1), the position could be calculated,
St=2.5s=5+15(2.5)+2(2.5)2=55maway from the signpost
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