Test: Acceleration (NCERT) - NEET MCQ

As x ∝ t³

velocity, v  ∝ 3t²

Acceleration, a ∝ 6t 

The slope of the tangent drown on a velocity-time graph at any instant of time is equal to the instantaneous acceleration.

The vertical axis will represent the acceleration of the object. The slope of the acceleration graph will represent a quantity called the jerk. This jerk is the rate of change of the acceleration. The area under this acceleration graph represents the change in velocity. Also, this area under the acceleration-time graph for some time interval will be the change in velocity during that time interval. Multiplying this acceleration by the time interval will be equivalent to finding the area under the curve.

For uniform motion with zero acceleration, v-t graph is a straight line parallel to the time axis.

Given: 

Squaring both sides, we get, v²=49+x
Differentiating both sides w.r.t. t, we get, 

Given: x² = at² + b
Differentiating w.r.t t on both sides, we get

Again differentiating w.r.t on both sides, we get

Let the car be accelerated from A to B. It moves with uniform velocity from B to C and then moves with uniform declaration from C to D.
For the motion of car from A to B,

For the motion of car from B to C

S = 9km = 9000m

For the motion of car from C to D,

Total time taken = t₁ + t₂ + t₃ ;

= 50s + 900s + 100s = 1050s

Let u be initial velocity and a be uniform acceleration.

Average velocities in the intervals from 0 to t₁, t₁ to t₂ and t₂ to t₃ are


Subtract (i) from (ii), we get

Subtract (ii) from (iii), we get

Divide (iv) by (v), we get

Given: a = p − qx
At maximum velocity, a = 0
⇒ 0 = p - qx or x = p/q
∴  Velocity is maximum at  x = p/q
since Acceleration 


vdv = (p − qx)dx 
Integrating both sides of the above equation, we get

x = t - sint;
v = dx/dt = 1 - cost;
a = dv/dt = sint
∴ x(t) > 0  for all values of t > 0 and v(t) can be zero for one value of t. a(t) can zero for one value of t.

For zero acceleration, the position-time graph is a straight line.

For the given velocity-displacement graph, intercept = v₀ and slope = -v₀/x₀

Thus, the equation of given line of velocity-displacement graph is

It is a straight line with positive slope and a negative intercept.

The variation of a with x is as shown in the above figure.

in the given interval the slop of v-t graph (i.e acceleration) is neither constant nor uniform. Therefore relations (a), (b) and (c) are incorrect but (d) is correct.