Question

An auto travelling along a straight road increases its speed from 30.0 m s^-1 to 50.0 m s^-1 in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?

• A. 6.0 s
• B. 4.5 s
• C. 3.6 s
• D. 7.0 s

Answer 1

Answer: (B)

4.5 s

Answer 2

Lat a be constant accelerations of auto

Here, u $30 \mathrm {~ms} ^ { - 1 }$ v = $50 \mathrm {~ms} ^ { - 1 }$ , S = 180 m

As $\mathrm { V } ^ { 2 } - \mathrm { u } ^ { 2 } = 2 \mathrm { aS }$

$( 50 ) ^ { 2 } - ( 30 ) ^ { 2 } = 2 \times \mathrm { a } \times 180$

(2500)-(900) = $2 \times a \times 180$

$\mathrm { a } = \frac { 1600 } { 2 \times 180 } = \frac { 40 } { 9 } \mathrm {~ms} ^ { - 2 }$

As $S = u t + \frac { 1 } { 2 } a t ^ { 2 }$

$180 = 30 \times \mathrm { t } + \frac { 1 } { 2 } \times \frac { 40 } { 9 } \times \mathrm { t } ^ { 2 }$

$180 = 30 t + \frac { 20 } { 9 } t ^ { 2 }$

$18 = 3 \mathrm { t } + \frac { 2 } { 9 } \mathrm { t } ^ { 2 }$

$\frac { 2 } { 9 } t ^ { 2 } + 3 t - 18 = 0$

$2 t ^ { 2 } + 27 t - 162 = 0$

Solving this quadratic by quadratic formula, we get

$r = \frac { - 27 \pm \sqrt { ( 27 ) ^ { 2 } - 4 ( 2 ) ( - 162 ) } } { 4 }$ $= \frac { - 27 \pm 45 } { 4 } = 4.5 \mathrm {~s} , - 18 \mathrm {~s}$

t can’t be negative

$\therefore \mathrm { t } = 4.5 \mathrm {~s}$