Home
Class 11
PHYSICS
A body starts at t=0 with velcoity u and...

A body starts at `t=0` with velcoity `u` and travels along a straight linge. The body has a constant acceleration(a). Draw the acceleration-time graph from `t=0` to `t=10 s` for the following cases:
a. `u=8 ms^(-1)`, a=2s^(-2)`
b. `u=8 ms^(-1)`, a=-2 ms^(-2)`
c. `u=- 8ms^(-1)` ,a=2 ms^(-2)`
d. `u=-8ms^(-1)` ,a=-2 ms^(-2)`.

Text Solution

Verified by Experts

a. `a=2 ms^(-2), u=8 ms^(1)`
`v=u+at=8+2t, S=ut+(1)/(2)a t^(2) =8t +t^(2)`
.
b. `a=-2 ms^(-2), u=8ms^(-1)`
`v=u+at=-2t, S=ut+(1)/(2) at^(2) =8t-t^(2)`
.
c. `a=2 ms^(-2),u=-8 ms^(-1)`
`v=u+at=-82t, S=ut +(1)/(2) at^(2) =- 8t+t^(2)`
.
d. `a=-2 ms^(-2) ,u=-8 ms^(-1)`
`v=u+at =-8-2t, S=ut+(1)/(2) at^(2)=-8t-t^(2)`
.
Promotional Banner

Topper's Solved these Questions

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Subjective|29 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Single Correct|52 Videos

  • KINEMATICS-1

    CENGAGE PHYSICS|Exercise Exercise 4.3|15 Videos

  • GRAVITATION

    CENGAGE PHYSICS|Exercise INTEGER_TYPE|1 Videos

  • KINEMATICS-2

    CENGAGE PHYSICS|Exercise Exercise Integer|9 Videos

Similar Questions

Explore conceptually related problems

A particle starts moving with initial velocity u=25ms^-1 and retardation a=-2ms^-2 . Draw the velocity-time graph.

A body starts with an initial velocity of 10ms-1 and is moving along a straight line with constant acceleration. When the velocity of the particle is 50ms- the acceleration is reversed in direction. The speed of the particle when it again reaches the starting point is

At time t = 0, a car moving along a straight line has a velocity of 16 ms^-1. It slows down with an acceleration of -0.5t ms^-2, where t is in second. Mark the correct statement (s).

A body of mass 2kg travels according to the law x (t) = pt + qt^(2) + rt^(3) where p = 3ms^(-1),q = 4 ms^(-2) and r = 5ms^(-3) .

The velocity- time graph of the particle moving along a straight line is shown. The rate of acceleration and deceleration is constant and it is equal to 5 ms^(-2) . If the average velocity during the motion is 20 ms^(-1) , then the value of t is

A particle starts from rest at t=0 and moves along a straight line with a variable acceleration given by a (t)=(3t+1)ms^(-2) ,where t],[ is in seconds.The velocity of the particle at t=4s is

The velocity of a particle are given by (4t – 2)" ms"^(–1) along x–axis. Calculate the average acceleration of particle during the time interval from t = 1 to t = 2s.

The acceleration time graph of a particle is shown in the Fig. 2 (CF). 11. At time t= 10 s is the particle is 8 ms^(-1) . Its velocity t= 10 s is. .

A body starting from rest has an acceleration of 4ms^(-2) . Calculate distnce travelled by it in 5th second.

Two cars (A) and (B) travel in straight line . The distance of (A) from the starting point is given as a function of time be a_A (t) = pt + qt^2 , with p = 2 . 60 ms^(-1) and q= 1.20 ms^(-2) . The distance of (B) from the starting pint is x_B (t) = rt^2 - st^3 are r= 2. 80 ms^(-2) and s = 0.20 ms^(-3) . Answer the following questions , At what time (s) do the cars (A) and (B) have the same acceleration ?