Question

A car accelerates from 6ms −1 to 16ms^-1 in 10 sec. Calculate (a) the acceleration and (b) the distance covered by the car in that time.

Answer 1

Explanation:

here u=6m/s,v=16m/s,t=10sec

now v=u+at

a=v-u/t

a=16-6/10

a=1m/s²

s=ut+1/2at²

s=6×10+1/2×1×100

s=60+50

s=110m

Answer 2

Answer :-

a) 1m/s^2

Explanation :-

Given :

Initial velocity = 6m/s

Final velocity = 16m/s

Time taken = 10s

To Find :

Distance travelled = ?

Acceleration = ?

Solution :

We know,

$\boxed{\sf{}a=\dfrac{v-u}{t}}$

Here,

a is the accelration,

v is the final velocity,

u is the initial velocity and,

t is the time taken.

Put their values and find “a”

$\sf{}:\implies a=\dfrac{16-6}{10}$

$\sf{}:\implies a=\dfrac{10}{10}$

$\sf{}\therefore a=1m/s^2$

Therefore,acceleration is equal to 2m/s^2

According to the second equation of motion,

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

here,

s is the distance travelled,

u is the initial velocity,

a is the acceleration and,

t is the time.

Put their values and find “s”

$:\implies \sf{}s=6\times 10+\dfrac{1}{2}\times1\times10^2$

$\sf{}:\implies s=60+\dfrac{1}{2}\times1\times100$

$\sf{}:\implies s =60+50$

$\sf{}\therefore s=110m$

Therefore,distance travelled is equal to 110m