The distance travelled by a motor car in t seconds after the brakes are applied is s feet, where s=22t−12t2 The distance travelled by the car before it stops, is

A
$10.08 f t$
B
$10 f t$
C
$11 f t$
D
$11.5 f t$

Solution:

We have,

$s = 22 t - 12 t^{2} \Rightarrow \frac{d s}{d t} = 22 - 24 t$

When the car stops, $\frac{d s}{d t} = 0$

$\begin{array}{l} ∴ 22 - 24 t = 0 \Rightarrow t = \frac{11}{22} \\ ∴ s = 22 t - 12 t^{2} \Rightarrow s = 22 \times \frac{11}{22} - 12 \times {(\frac{11}{22})}^{2} = 10.08 ft . \end{array}$

The distance travelled by a motor car in $t$ seconds after the brakes are applied is $s$ feet, where $s = 22 t - 12 t^{2}$ The distance travelled by the
car before it stops, is

Solution:

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