For a race a distance of 224 meters can be covered by P in 28 seconds and Q in 32 seconds. By what distance does P defeat Q eventually?
Answers
→Speed of P= 224/28 = 8m/s
→Speed of Q=224/32 = 7m/s
➾Difference = 4 seconds
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=》Therefore, Distance covered by P in time= 8m/s x 4 seconds = 32 metres
Answer:
Explanation:
If car B travels 224 m in 32 seconds, how far will it have gone in 28 seconds?
Depends on what we assume about its acceleration, and initial velocity.
With no acceleration, the velocity is a constant 7 m/s, so it has travelled v*t = 7 * 28 = 196 m, and so it's 28 m behind A.
With constant forward acceleration and zero initial velocity, x = 1/2 A*t^2, so A = .4375 m/s^2 (to get to the finish line on time) At 28 seconds it has traveled 171.5 m, and is 52.5 m behind.
It could also be constantly braking at .4375 m/s^2 from an initial velocity of 14 m/s and come to a stop at the finish in 32 seconds, so at 28 seconds would have travelled v*t + 1/2 A * t^2 = 14*28 + 1/2*-.4375*28^2 = 220.5 m, and be only 3.5 m behind. (Though I'm not sure why the driver would be braking during a drag race!)