Question

A car moving at 20 m/s slows down at -1.5 m/s^2 to a velocity of 10 m/s. How far did the car go during the slow down? Thank you so much.

Answer 1

IS that meant to be delta

Answer 2

\[20^{2}= 10^{2} + 2(-1.5)(distance)\]

Answer 3

use \[{v_f}^2={v_0}^2+2ad\]
\[\frac{400-100}{-3}=d=-100\]

Answer 4

v f 2 =v 0 2 +2ad
400−100 −3 =d=−100
or
20 2 =10 2 +2(−1.5)(distance)

Answer 5

Given, initial velocity, u = 20m/s
velocity at particular time, v = 10m/s
acceleration, a = -1.5 m/s^2

From equation of motion,
\[v ^{2}-u ^{2}=2aS\]where S is the distance travelled
therefore,
\[(10)^{2}-(20)^{2}=2\times(-1.5)\times S\]100 - 400 = -3 S
-300 = -3 S
S=100 meter