How Long Will It Take for a Car to Accelerate from 55 km/h to 75 km/h?

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To determine the time it takes for a car to accelerate from 55 km/h to 75 km/h, one must consider the constant power assumption and the relationship between kinetic energy and speed. The initial calculations using acceleration and distance were flawed, as they did not account for the dynamics of power and energy. It was emphasized that since power remains constant, the time to increase speed by 20 km/h will be longer than the previous acceleration from 35 km/h to 55 km/h. The correct approach involves calculating the work done and using the power equation, rather than simply applying the basic kinematic equations. Ultimately, the discussion highlighted the need to convert speeds to m/s for accurate calculations.
Sportsman4920
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If a 1500 kg car can accelerate from 35 km/h to 55 km/h in 3.8 s, how long will it take to accelerate from 55 km/h to 75 km/h? Assume the power stays the same, and neglect frictional losses

what I have tried so far:
55=35+a3.8

a=5.263 km/h squared

75 squared= 55 squared + 2(5.263squared)d

d=46.93

46.93=55+2.6315t
-8.067= 2.6315t

t=3.06 but the answer is obviously wrong
 
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Since acceleration is constant, you can use v(t) = v_{0} + at. Simply apply that formula to the case when v0 = 55 km/h, and v(t) = 75.
 
diodnt they ask you how LONG it takes it takes to reach 75km/h from 55km/h? That means time

you have the initial velocity, you have the final velocity, you have your acceleration, can't you find the time WITHOUT having to find distance?
 
radou said:
Since acceleration is constant, you can use v(t) = v_{0} + at. Simply apply that formula to the case when v0 = 55 km/h, and v(t) = 75.

The question says assume power is constant; since kinetic energy has v^2 in it, it will take longer to up the speed by 20km/h than previously.
 
BerryBoy said:
The question says assume power is constant; since kinetic energy has v^2 in it, it will take longer to up the speed by 20km/h than previously.

You're right. I missed the word 'power'. So it becomes a dynamics problem.
 
Last edited:
when I plug the numbers into the equation 75=55+at

I get 20=5.263t

t=3.8001, but when I plug that in it is wrong. And that would be the same time as it took to get from 35 to 55
??
 
Sportsman4920 said:
when I plug the numbers into the equation 75=55+at

I get 20=5.263t

t=3.8001, but when I plug that in it is wrong. And that would be the same time as it took to get from 35 to 55
??

Again, read BerryBoy's post. As stated, it's a dynamics problem. You have to calculate changes of kinetic energy, which equal the work done. Then plug that work into the equation for power, which, as said, remains constant.

P.S. Convert [km/h] to [m/s].
 
Last edited:

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