A scooter starts from rest moves in a straight line with a constant acceleration and covers a distan

English

Gujarati

Hindi

7. MOTION

medium

A scooter starts from rest moves in a straight line with a constant acceleration and covers a distance of $64 \,m$ in $4 \,s$

$(i)$ Calculate its acceleration and its final velocity.

$(ii)$ At what time the scooter had covered half the total distance ?

Option A

Option B

Option C

Option D

Solution

$(i)$ $a=? ; u=0 ; S =64 m ; t=4 s$

Using the equation $S=u t+1 / 2 a t^{2},$ we have

$64=0+a(4)^{2}$

or $a=4 m s ^{-2}$

Using the equation $v=u+\mu t,$ we have

$v=0+4 \times 4=16 m s ^{-1}$

$(ii)$ Given $S=32 m , t=?, d=4 m s ^{-2}$

Using the equation $S=u t+\frac{1}{2} a t^{2}$

$32=0^{\prime}+4 \times(t)^{2}$ or $t=2.82 s$

Standard 9

Science

Share

Similar Questions

Answer the following questions

$(i)$ State the type of motion shown by a freely falling stone.

$(ii)$ When a stone is thrown vertically upwards its velocity is continuously decreased. Why ?

$(iii)$ Give an example of a motion in which average velocity is zero, but the average speed is not zero.

easy

A frog hops along a straight line path from point $'A^{\prime}$ to point ${ }^{\prime} B ^{\prime}$ in $10\, s$ and then turns and hops to point ${ }^{\prime} C^{\prime}$ in another $5\, s$. Calculate the average speed and average velocity of the frog for the motion between $(a)(A)$ to $(B)(b)(A)$ to $(C)($ through $B)$

medium

Area under velocity$-$time graph is equal to the

easy

There is an argument about uniform acceleration between $Mr$ $X$ and $Mr$ $Y.$ $Mr$ $X$ says "acceleration means that farther you go faster you go". $Mr$ $Y$ says "acceleration means that longer you go the faster you go". Whose statement is correct ?

medium

What do you understand by the displacement$-$time graph ? Draw a displacement-time graph for a girl going to school with uniform velocity. How can we calculate the uniform velocity from it ?

medium