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What is the final velocity of a 1400 kg drag racer that’s engine applies 910000 N of Force for 120m
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Answers ( )
Answer:
468,000,000 m/s
Explanation:
According to newtons first law of motion;
Force = mass * acceleration
F = ma
where a = v-u/t
v is the final velocity
u is the initial velocity
t is the time taken
F = m(v-u/t)
Ft = m(v-u)
Given parameters;
F = 910,000N
t = 120m = 120 *60 = 7200secs
m = 1400kg
u = 0m/s
Required parameter
Final velocity v
Substituting the given parameters into the formula to get v;
Ft = m(v-u)
910,000 * 72000 = 1400(v-0)
910,000 * 72000 = 1400v
v = (910,000 * 72000)/1400
v = 910,000 * 720/14
v = 910,000 * 60
v = 468,000,000 m/s
Hence the final velocity of the drag racer is 468,000,000 m/s