A car starts from rest and moves in a straight line with constant acceleration $$\(a \mathrm{~m} \mathrm{~s}^{-2}\)$$ for a distance of $$\(50 \mathrm{~m}\)$$. The car then travels with constant velocity for $$\(500 \mathrm{~m}\)$$ for a period of $$\(25 \mathrm{~s}\)$$, before decelerating to rest. The magnitude of this deceleration is $$\(2 a \mathrm{~m} \mathrm{~s}^{-2}\)$$. Find the total time for which the car is in motion. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No:9709_s20_qp_43 Year:2020 Question No:4(c)
Answer:
Time to accelerate \(=20 / 4=5 \mathrm{~s}\)
Deceleration time \(=2.5 \mathrm{~s}\)
So total time \(=5+25+2.5=32.5 \mathrm{~s}\)
Deceleration time \(=2.5 \mathrm{~s}\)
So total time \(=5+25+2.5=32.5 \mathrm{~s}\)
Knowledge points:
4.2.2.1 the area under a velocity–time graph represents displacement
4.2.2.2 the gradient of a displacement–time graph represents velocity
4.2.2.3 the gradient of a velocity–time graph represents acceleration
Solution:
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