If a 1500 \rm kg car can accelerate from 35 km/h to 50 km/h in 3.3 s, how long will it take to accelerate from 50 km/h to 70 km/h? Assume the power stays the same, and neglect frictional losses.
4.4 is not correct
4.4 is not correct
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Since power is constant, note that acceleration is not constant.
(You can work out the details for yourself by if you want - it's because constant power would result in the change in kinetic energy being proportional to t, but constant acceleration results in the change of kinetic energy being proportional to t², assuming no friction.)
35km/h = 35 x 1000 / 3600 m/s = 9.722m/s
50km/h = 50 x 1000 / 3600 m/s = 13.89m/s
Kinetic energy at 35km/h = ½ x 1500 x 9.722² = 70888J
Kinetic energy at 50km/h = ½ x 1500 x 13.89² = 144699J
Power = change in kinetic energy / time (P = E/t)
= (144699-70888) / 3.3
= 22367W
70km/h = 70 x 1000 / 3600 m/s = 19.44m/s (or just double the value for 35km/h)
Kinetic energy at 70km/h = ½ x 1500 x 19.44² = 283435J (or just multiply the KE at 35km/h by 4)
Change in kinetic energy from 50km/h to 70km/h = 283435 - 144699 = 138736J
Since P = E/t, t = E/P
t = 138736/22367 = 6.2s
(You can work out the details for yourself by if you want - it's because constant power would result in the change in kinetic energy being proportional to t, but constant acceleration results in the change of kinetic energy being proportional to t², assuming no friction.)
35km/h = 35 x 1000 / 3600 m/s = 9.722m/s
50km/h = 50 x 1000 / 3600 m/s = 13.89m/s
Kinetic energy at 35km/h = ½ x 1500 x 9.722² = 70888J
Kinetic energy at 50km/h = ½ x 1500 x 13.89² = 144699J
Power = change in kinetic energy / time (P = E/t)
= (144699-70888) / 3.3
= 22367W
70km/h = 70 x 1000 / 3600 m/s = 19.44m/s (or just double the value for 35km/h)
Kinetic energy at 70km/h = ½ x 1500 x 19.44² = 283435J (or just multiply the KE at 35km/h by 4)
Change in kinetic energy from 50km/h to 70km/h = 283435 - 144699 = 138736J
Since P = E/t, t = E/P
t = 138736/22367 = 6.2s
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Yes it is.