Math

QuestionA car stops from 32 m/s32 \mathrm{~m/s} with an acceleration of 5 m/s2-5 \mathrm{~m/s}^{2}. How far did it travel before stopping?

Studdy Solution

STEP 1

Assumptions1. The initial velocity of the car is 32 m/s32 \mathrm{~m/s} . The final velocity of the car is 0 m/s0 \mathrm{~m/s} (since the car comes to a stop)
3. The acceleration of the car is -5 \mathrm{~m/s^} (negative because the car is decelerating)
4. We need to find the distance travelled by the car before it stopped

STEP 2

We can use the equation of motion to solve for the distance travelled by the car. The equation of motion is given byv2=u2+2asv^2 = u^2 +2aswhere- vv is the final velocity- uu is the initial velocity- aa is the acceleration- ss is the distance travelled

STEP 3

Now, plug in the given values for the initial velocity, final velocity, and acceleration to solve for the distance.
02=322+2(5)s0^2 =32^2 +2(-5)s

STEP 4

implify the equation.
0=102410s0 =1024 -10s

STEP 5

Rearrange the equation to solve for ss.
10s=102410s =1024

STEP 6

Finally, divide both sides of the equation by10 to solve for ss.
s=102410s = \frac{1024}{10}

STEP 7

Calculate the distance travelled by the car before it stopped.
s=102410=102.4 ms = \frac{1024}{10} =102.4 \mathrm{~m}The car travelled102.4 meters before it stopped.

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