Question

If you double the speed of a vehicle how much does the force in a crash increase by?

Answer 1

The force that gets generated when the object collides is called the force of impact. The faster the vehicle moves, the greater will be the striking power of the vehicle. This force of impact increases in proportion to the square of the increase in speed.
Formula used:
$I=Ft$

Complete answer:
The force of impact increases in proportion to the square of the increase in speed. Thus, if the speed of the vehicle is doubled, then, the force in a crash or the force of impact increases four times. Similarly, if the speed of the vehicle is tripled, then, the force in a crash or the force of impact increases nine times.
Impulse is the product of the force applied to an object/body and the amount of time applied.
$I=Ft$
Where I is the impulse, F is the force applied and t is the time invested.

& F=\dfrac{I}{t} \\\ & \therefore F=\dfrac{mv}{t} \\\ \end{aligned}$$ Where m is the mass of the body and v is its velocity. The force applied and the time invested is inversely proportional to each other. If the force is increased, then, the time gets reduced by an equal amount. (Because, the faster the velocity, the lesser will be the time taken to crash). The condition is to double the velocity. So, we get, $$\begin{aligned} & F=\dfrac{m(2v)}{t} \\\ & \Rightarrow F=\dfrac{m(2v)}{{}^{t}/{}_{2}} \\\ & \therefore F=4\times \dfrac{mv}{t} \\\ \end{aligned}$$ Thus, we have mathematically proved that the force of crash increases by four times, if the velocity of the vehicle is doubled. $$\therefore $$ If we double the speed of a vehicle, then, by four times the force in a crash increases. **Note:** When 2 vehicles having different weights moving at the same rate of speed are involved in a crash, then, the vehicle that weight lighter will experience the greater force of impact, thus, will suffer from a greater deceleration and may even get pushed back.