Question

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

Answer 1

Answer:

Acceleration = 1m/s/^2,  Distance = 110m

Explanation:

Since, acceleration is v-u/t

Acceleration is 16-6/10 = 1m/s^2

Using (2)nd equation of motion,

s= ut + 1/2at^2

s = 6*10 + 1/2*1*(10)^2

s = 60 + 50

s = 110m

So, distance travelled is 110m.

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Answer 2

Given :

▪ Initial velocity = 6mps

▪ Final velocity = 16mps

▪ Time interval = 10s

To Find :

▪ Acceleration of car.

▪ Distance covered by car in the given interval of time.

Concept :

✒ Acceleration is defined as the ratio of change in velocity to the time interval.

✒ It is a vector quantity.

✒ It can be positive, negative and zero.

✒ Since, acceeration has said to be constant throughout the motion, we can apply equation of kinematics to solve the second question.

Calculation :

✳ Acceleration of car :

$\implies\sf\:a=\dfrac{v-u}{t}\\ \\ \implies\sf\:a=\dfrac{16-6}{10}\\ \\ \implies\sf\:a=\dfrac{10}{10}\\ \\ \implies\underline{\boxed{\bf{\pink{a=1\:ms^{-2}}}}}$

✴ Distance covered by car :

$\implies\sf\:v^2-u^2=2as\\ \\ \implies\sf\:(16)^2-(6)^2=2(1)(s)\\ \\ \implies\sf\:\sf\:256-36=2s\\ \\ \implies\sf\:220=2s\\ \\ \implies\underline{\boxed{\bf{\purple{s=110m}}}}$