Question

A car accelerates from $6m{s}^{-1}$ to $16m{s}^{-1}$ in $10s$. Calculate:

i. The acceleration

ii. The distance covered by car in that time?

Answer 1

Step 1: Given data:

Initial velocity of the car($u$) = $6m{s}^{-1}$

The final velocity of the car($v$) = $16m{s}^{-1}$

Time($t$) taken = $10s$

Step 2: Formula used:

The first and second equations of motion are given as;

$v=u+at$
And

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

Where, $v$ = final velocity

$u$ = Initial velocity

$a$ = Acceleration in the body

$t$ = Time taken by the body

$s$ = Distance covered

Step 3: i) Calculation of the acceleration in the car:

Using the first equation of motion,

$16m{s}^{-1}=6m{s}^{-1}+a\times 10s$

$\Rightarrow 6+10a=16$

$\Rightarrow 10a=10$

$\Rightarrow a=1m{s}^{-2}$

Step 4: ii) Calculation of the distance covered by the car:

Using the second equation of motion, we get

$s=6m{s}^{-1}\times 10s+\frac{1}{2}\times 1m{s}^{-2}\times {\left(10s\right)}^2$

$s=60+50$

$s=110m$

Thus,

i. The acceleration = $1m{s}^{-2}$

ii. The distance covered by car in that time = $110m$