Question

If u = 9, a = 4, s = 35, then find the value of t

Answer 1

• initial velocity=u=9m/s
• Acceleration=a=4m/s^2
• Distance=s=35m
• Time=t=?

ACCORDING TO 2ND EQUATION OF KINEMATICS

$\boxed{\sf s=ut+\dfrac{1}{2}at^2}$

• putting values

$\\ \sf\longmapsto 35=9t+\dfrac{1}{2}\times 4t^2$

$\\ \sf\longmapsto 35=9t+2t^2$

$\\ \sf\longmapsto 2t^2+9t-35=0$

$\\ \sf\longmapsto 2t^2+14t-5t-35=0$

$\\ \sf\longmapsto 2t(t+7)-5(t+7)=0$

$\\ \sf\longmapsto (2t-5)(t+7)=0$

$\\ \sf\longmapsto (2t-5)=0\:or\:(t+7)=0$

$\\ \sf\longmapsto t=\dfrac{5}{2}\;or\:t=-7$

• Time can't be negative

$\\ \sf\longmapsto t=\dfrac{5}{2}$

$\\ \sf\longmapsto t=2.5s$

Answer 2

$\huge\bf\red{\mid{\fbox{\underline{\maltese{Answer}}\maltese}}}$

Given :-

$\blue\longrightarrow \: u = 9m/s \: \: \: \: \: \: \: \\ \blue\longrightarrow \:a = 4m/s² \: \: \: \: \: \: \: \\ \blue\longrightarrow \:s = 35 \: m \: \: \: \: \: \: \: \: \: \:$

To Find :-

$\red\longrightarrow \: t = \: ?$

Formula Used :-

$\pink\longrightarrow \: s = ut + \frac{1}{2} \: a {t}^{2}$

Solution :-

$\green\longrightarrow \: 35 \: m = 9 \times t + \frac{1}{2} \times 4 \times {t}^{2} \\ \green\longrightarrow \: 35 \: m = 9t + \frac{1}{\cancel{2}} \times \cancel{4} \times {t}^{2} \: \: \: \: </u><u> \\ \green\longrightarrow 35 \: m = 9t + 2 {t}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \green\longrightarrow \: 35 \: m + 9t - 2 {t}^{2} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \\ \green\longrightarrow \: 35 + 14t - 5t - 2 {t}^{2} = 0 \: \: \: \: \: \\ \green\longrightarrow \: 2t(7 + t) - 5(7 + t) = 0 \: \: \: \\ \green\longrightarrow \: (2t - 5)(7 + t) = 0 \: \: \: \: \: \: \: \: \: \: \: \:$

Therefore,

$\longrightarrow \: t = \frac{5}{2} \: or \: t = - 7$

We know that, time can never be in negative.

$\longrightarrow \: t = \frac{5}{2} = 2.5 \: seconds$

Therefore, time ( t) is 2.5 seconds.

Thank you!