This is a motion diagram of an object moving along the x-direction with constant

Question

This is a motion diagram of an object moving along the x-direction with constant acceleration. The dots 1, 2, 3.... Show the position of the object at equal time intervals delta t Which of the following a-1 graphs best matches the motion shown in the motion diagram? An object moves along the x-axis with constant acceleration. The initial position x2 is positive , the initial velocity is negative, and the acceleration is positive. which of the following V2-1 graphs best describe this motion? A motorcyclist heading east acceleration at a constant 4m/s2. At t=0 he is 5. 0 m east of a signpost, moving east at 15m/s. Find the position and velocity at t= 2. 0s. Where is he when his velocity is 25 m/s?

Explanation / Answer

Q>3..

In the diagram the dot increases its relative position with time..

when velocity increses the relative placing of the object at different time intervals also start increasing and when there in acceleration (constant or increasing), velocity increases.

In the diagram the relative spacing is incresing uniformly hence the acceleration is uniform.

Ans) A

Q>4

It is said that the body is having negative velocity and positive acceleration in the X-direction which means first the body is moving in negative X-direation and constantly increasing its velocity along positive X, which inturn includes first reduction along negative X then 0 and finally increasing along positive X. Hence the answer is B

Ans)B

Q>3'

Given,

int velocity u = 15m/s

accl a = 4m/s2

S int = 5m

a) t=2 sec

S=ut+at2/2

=15(2)+4(2)2/2 = 30+8 = 38m

final position = 38 + 5 = 43 m

b) V final = 25m/s

v2-u2=2as

(25)2-(15)2=2(4)(s)

625-225=8s

s=50m

final position = 50 + 5 = 55m

Q>4'

a) Given,

a=3m/s2

u=0m/s

let time be t

distance travelled by motor cyclist and police is same

dist travelled by police = at2/2 = 3t2/2

dist traveled by motor cyclist = ut = 15t

3t2/2 = 15t

t = 10 sec

b) t = 10s

v=at=3(10) = 30 m/s

c) s= vt = 15(10) = 150 m

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