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A lorry moving under uniform acceleration covers a distance of $180 m$ from the point $A$ to point $B$ along a straight path in $15 s$ . Its velocity, while passing through point $B,$ is ${20 ms}^{- 1}$ . What is its acceleration?

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1.07 m s^{- 2}

2.08 m s^{- 2}

1 m s^{- 2}

3.08 m s^{- 2}

answer is A.

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Detailed Solution

The acceleration of the lorry is

1.07 m s^{- 2}

.
Given data in the problem,
Calculation for velocity of the lorry
Distance covered by the lorry,

s = 180 m

Time,

t = 15 s

Final velocity,

v = 20 m s^{- 1}

So, from below formula, we can find the average velocity,
Average velocity,

v_{av} = \frac{Total distance travelled}{Total time taken}

\Rightarrow v_{av} = \frac{u + v}{2}

\Rightarrow v_{av} = \frac{s}{t}

\Rightarrow v_{av} = \frac{u + 20}{2}

\Rightarrow v_{av} = \frac{180}{15}

{\Rightarrow v}_{av} = 12

By using the first equation of motion,

v = u + at

Where,

v

= final velocity

u

= initial velocity

a

= acceleration

t

= time
Putting values in the above equation,

\Rightarrow u + 20 = 2 \times 12

\Rightarrow u = 24 - 20

\Rightarrow u = 4 m s^{- 1}

Calculation of acceleration,
Again, using the first equation of motion,

v = u + at

Where,

v

= final velocity

u

= initial velocity

a

= acceleration

t

= time
Rearranging the above equation,
Acceleration,

a = \frac{v - u}{t}

\Rightarrow a = \frac{20 - 4}{15}

\Rightarrow a = \frac{16}{15}

\Rightarrow a = 1.07 m s^{- 2}

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