JEE Main 2017 · 8 Apr · Shift 1 · Physics Q2

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PhysicsSingle correct+4 / -1Question 2 of 90

Which graph corresponds to an object moving with a constant negative acceleration and a positive velocity ?

Velocity against time: a curve falling to a minimum and rising again

Velocity against time: a U-shaped curve, falling to a minimum and rising again.

Velocity against time: a curve rising more and more steeply

Velocity against time: starting flat and rising more and more steeply.

Velocity against distance: a curve falling more and more steeply to zero (concave down)

Velocity against distance: starting at a maximum and falling ever more steeply to zero on the distance axis (a quarter-ellipse shape).

Velocity against distance: a curve falling steeply and flattening out (concave up)

Velocity against distance: falling steeply at first and flattening out towards the distance axis.

Answer and worked solution

Official answer: C

With constant acceleration $a<0$ , $v^2=v_0^2+2ax$ : velocity against distance is a downward-opening parabola, falling ever more steeply until $v=0$ . That is graph (3). Against time the graph would be a straight line, so (1) and (2) are ruled out.

Topics:Motion graphs Motion in a straight line

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JEE Main 2017 · 8 Apr · Shift 1, Physics question 2