Test: Instantaneous Velocity & Speed , Acceleration (April 17) - JEE MCQ

Speedometer measures the speed of the car in km h⁻¹.

Here, x = a + bt²

Where, a = 8.5m and b = 2.5ms⁻²

At t = 2s,
v = 2(2.5ms⁻²)(2s) = 10ms⁻¹

Given: x = 9t² − t³

for Maximum speed, dv/dt
= 0 ⇒ 18 - 6t = 0
∴ t = 3s.
∴ x_max = 81m - 27m
= 54 m. (From x = 9t² - t³)

As OP < OQ, A lives closer to the school than B.

(b) For x = 0, t = 0 for A; while t has some finite value for B. Therefore,

A starts from the school earlier than B.

(c) Slope of x - t graph is equal to the speed. Since the slope of x-t for B is greater than that for A. Therefore, B walks faster than A.

(d) Corresponding to points P and Q the value of t from x - t graphs for A and B is same. Thus, both A and B reach home at the same time.

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.

Given: 

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

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

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

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.