A tram starts from rest and moves with uniform acceleration for $$\(20 \mathrm{~s}\)$$. The tram then travels at a constant speed, $$\(V \mathrm{~m} \mathrm{~s}^{-1}\)$$, for $$\(170 \mathrm{~s}\)$$ before being brought to rest with a uniform deceleration of magnitude twice that of the acceleration. The total distance travelled by the tram is $$\(2.775 \mathrm{~km}\)$$. Find $$\(V\)$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No:9709_s20_qp_42 Year:2020 Question No:1(b)
Answer:
\(0.5(170+200) v=2775\)
\(v=15\)
\(v=15\)
Knowledge points:
4.2.2.1 the area under a velocity–time graph represents displacement
Solution:
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